Cls Jeead-14-15 Xi Mat Target-4 Set-1 Chapter-16
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Chapter
16
Probability Solutions SECTION - A School/Board Exam. Type Questions Very Short Answer Type Questions : 1.
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
2.
Total number of outcomes = 100 Favourable outcomes = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90}
Number of favourable outcomes = 10
Hence, required probability = 3.
10 1 100 10
Leap year consists of 52 weeks and 2 days. These 2 days can be {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
Total number of outcomes = 7
And number of favourable outcomes = 2 Thus, required probability =
4. 5.
2 7
1 221 S = {(R, B1), (R, B2), (R, B3), (B1, R), (B1, B2), (B1, B3), (B2, B1), (B2, B3), (B2, R), (B3, R), (B3, B1), (B3, B2),} 4
6.
C2 2 Required probability = 7 C2 7
7.
4! 2! 4! 1 8! 35
8.
Let A be the event “getting a total of atleast 5” P(A) = 1 – P ( A) = 1 – P(getting a total of atmost 4)
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Probability
(School/Board Exams.) Solutions
Favourable elementary events to A = (1, 1, 1) (1, 1, 2), (2, 1, 1), (1, 2, 1)
P ( A)
4 216
[∵ Total number of elementary events = 216]
Thus, P ( A) 1 P ( A) = 1 9.
4 212 53 216 216 54
A four digit number is formed at random using the numerals, 1, 2, 3 and 5 Total number of elementary events = 4! = 24 Favourable elementary events = Th 1
Hence, required probability =
H
T
U
2
3
1
= 1 2 3 1 = 6
[∵ Unit digit can be 5 only]
6 1 24 4
10. P(A) = 0.25, P(B) = 0.50 P(A B) = 0.14 P(A B) = P(A) + P(B) – P(A B) = 0.25 + 0.50 – 0.14 = 0.61 Now, P ( A B ) P ( A B ) 1 P ( A B ) 1 0.61 = 0.39 Short Answer Type Questions : 11. P(E1 E2) = P(E1) + P(E2) – P(E1 E2)
P(E E2) = p1 + p2 – p3
Now, P (E1 E2 ) P (E1 E 2 ) 1 P (E1 E2 ) = 1 – (p1 + p2 – p3) = 1 – p1 – p2 + p3
P (E1 E2 ) = 1 – p1 – p2 + p3
12. Total number of elementary events =
30C 3
Favourable elementary events = {(1, 2, 3), (2, 3, 4), (3, 4, 5), …… (28, 29, 30)}
Number of favourable outcomes = 28
Thus, required probability =
28 30
C3
1 145
13. Let E be event that all the three drawn balls are of same colour. Then number of elements in favour of E = 3C3 + 4C3 All the possible different ways drawing three balls = 7C3 3
Required probability =
C3 4C3 7
C3
1 7
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(School/Board Exams.) Solutions 13
14.
52
Probability
103
C5 C5
15. Let P(A B) = K – d, P(A) = K, P(B) = K + d, and 1 = P(A B) = K + 2d
1 K 2
d
P(A) = K, P(B) = K + d =
P(A) : P(B) = K :
K 1 2
K 1 2K : K 1 2
16. We know that the probability of occurrence of an event is always less than or equal to 1 and it is given that P(A B C) 0.55
0.55 P(A B C) 1
Let P(B C) = x
0.55 0.2 + 0.5 + 0.4 – 0.06 – 0.05 – x + 0.04 1
0.55 1.03 – x 1
–1 x –1.03 –0.55
0.03 x 0.48
P(B C) lies in the interval [0.03, 0.48]
17. Let E and H denote the events that the student will pass in English and Hindi examination respectively
P (E H ) 0.1
P (E H ) 0.1
1 – P(E H) = 0.1
P(EH) = 0.9
P(E) + P(H) – P(E H) = 0.9
0.75 + P(H) – 0.5 = 0.9
P(H) = 0.65
18. Since, the events are mutually exclusive, 1 3 p 1 p 1 2p 1 3 4 2
0
0 4 + 12p + 3 – 3p + 6 – 12p 12
– 13 – 3p – 1
1 13 p 3 3
Also 0
1 3p 1 2 1⇒ p 3 3 3
…(i)
…(ii)
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Probability
(School/Board Exams.) Solutions
0
1 2p 1 1 1⇒ p 2 2 2
…(iii)
0
1 p 1⇒ – 3 p 1 4
…(iv)
From (i), (ii), (iii) and (iv), we get 1 1 p 3 2
19. Total number of elementary events =
20C 1
= 20
A multiple of 3 or 7 can be obtained as follows: 3, 6, 9, 12, 15, 18, 7, 14 So, number of favourable elementary events = 8 Required probability =
20.
P ( A)
P (B )
20
C2
30
C2
22
C2
30
(A : Event of getting 2 apples)
(B : Event of getting 2 good items)
C2
P( A B)
8 2 20 5
15 30
C2 C2
i.e., P(getting 2 items which are good apples)
Now, P(A B) = P(A) + P(B) – P(A B) 20
=
30
C2 C2
22 30
C2 C2
15 30
C2 C2
316 435
21. The quadratic equation will have real roots p2 4
1 ( p 2) 0 4
p2 – p – 2 0
(p – 2) (p + 1) 0
p2
Required probability =
[i.e. b2 – 4ac 0]
(∵ p 0) 52 3 5 5
22. Let E be the event “sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3”
E : Event “sum of the numbers obtained on the two dice is a multiple of 2 or 3” Total number of elementary events = 6 6 = 36 Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
(School/Board Exams.) Solutions
Probability
105
Favourable elementary events to event E = {(1, 1), (1, 2), (1, 3), (1, 5,) (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (4, 6), (5, 1), (5, 3), (5, 4), (5, 5), (6, 2), (6, 3), (6, 4) (6, 6)}
P (E )
24 2 36 3
Hence, P(E) = 1 – P (E ) 1
2 1 3 3
23. Favourable cases: With difference 1 : (1, 2, 3), (2, 3, 4) …… (9, 10, 11) = 9 With difference 2 : (1, 3, 5), (2, 4, 6) …… (7, 9, 11) = 7 With difference 3 : (1, 4, 7), (2, 5, 8), (5, 8, 11) = 5 With difference 4 : (1, 5, 9), (2, 6, 10), (3, 7, 11) = 3 With difference 5 : (1, 6, 11) = 1
Probability =
25 11
C3
5 33
24. A, B and C mutually exclusive and exhaustive events
P(A) + P(B) + P(C) = 1
P ( A)
19 4 P ( A) 1 ⇒ P ( A) 4 19
1 1 5 ⎡ ⎤ ⎢∵ P (C ) 2 P (B ) 2 2 (P ( A))⎥ ⎣ ⎦
5 5 P ( A) P ( A) 1 2 4
25. Probability that out of four drawn balls 2 are white, one red and one black. 5
=
C2 7C1 8C1 20
C4
112 969
26. Let x S. Then either x P, x Q or x P, x Q, or x P, x Q, x P, x Q. Out of the above four cases, three cases are favourable to the events P Q = .
⎛ 3⎞ Required probability = ⎜ ⎟ ⎝ 4⎠
20
27. Total number of ways in which 8 persons can sit on a round table is (8 – 1)! = 7!
Total number of elementary events 7! Consider two named individuals as one person, then there will be 7 persons, who can sit on a round table is (7 – 1)! = 6! ways. Also, two named individuals can be seated together in 2 ways
Favourable number of elementary events = 6! 2 So, required probability =
6! 2 2 7! 7
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Probability
(School/Board Exams.) Solutions
28. A : Both cards are black B : Both cards are queens
P(A or B) = P(A) + P(B) – P(A B) 26
= 3
29. P(none white) =
52
C2 C2
C0 7C2 10
C2
4
C2
52
C2
2
C2
52
C2
7 15
30. P(A) = P(B) = 2P(C) 1 = P(A B C) = P(A) + P(B) + P(C) = 5P(C)
1 2 , P ( A ) P (B ) 5 5
P (C )
P (B C ) P (B ) P (C )
2 1 3 5 5 5
Long Answer Type Questions : 31. P(A B) = P(A) + P(B) – P(A B)
1 1 2 P( A B) 2 4 5
P( A B)
1 2 1 3 4 5 2 20
P ( A B ) P ( A) P ( A B )
13
32. (i)
(iv) 33. (i)
=
53 2 1 20 20 10
C1
13
C1
C2
C1 4C1 52
4
(iii)
1 3 4 20
52
4
(ii)
=
C2
C2
52
C2
13
C2
52
C2
13 52
C2 C2
13 52
C2 C2
13 52
C2 C2
4
13 52
C2 C2
For red and king, there are 2 favourable cases
Probability =
2 1 52 26
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(School/Board Exams.) Solutions
(ii)
4
35.
28 7 52 13
Probability = C3
15
5
(ii)
107
For red or a king, there are 28 favourable cases
34. (i)
Probability
C3
=
4 455
C1 6C1 4C1 15
C3
=
24 91
4 9
36. Total number of cases = 10 10 = 100 For real roots, b2 4c Number of possible ways is given in the following table:
Value of C
Value of 4c
Value of b
No. of ways of b
1
4
2, 3, 4, 5, 6, 7, 8, 9, 10
9
2
8
3, 4, 5, 6, … 10
8
3
12
4, 5, 6, … 10
7
4
16
4, 5, 6, … 10
7
5
20
5, 6, 7, … 10
6
6
24
5, 6, 7, … 10
6
7
28
6, 7, 8, 9, 10
5
8
32
6, 7, 8, 9, 10
5
9
36
6, 7, 8, 9, 10
5
10
40
7, 8, 9, 10
4 62
Probability =
62 0.62 100
37. Total number of cases = 100 E : Event that n +
100 50 n
n2 – 50n + 100 > 0
(n – 25)2 > 525
| n 25| 525
n 25 525
n 25 525
or
n 25 525
n 25 525
n 2 or n 48
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Probability
(School/Board Exams.) Solutions
E = {1, 2, 48, 49, 50, …… 100}
P (E )
55 0.55 100
38. ABCD is a trapezium in which AB and CD are parallel sides Since AB || CD Hence,
B + C = 180°
Now,
A + D = 180°
D
A D A D 90 ⇒ 90 2 2 2 2
A D tan cot 2 2
Similarly, tan
tan
B
A
B C cot 2 2
A B C D tan tan tan 1 2 2 2 2
The event tan
C
A B C D tan tan tan 1 is a certain event 2 2 2 2
Probability = 1
39. Total number of elementary events = Now,A = {2, 4, 6, ……50}
50C 1
= 50 P(A) =
25 50
B = {3, 6, 9, …… 48}
P(B) =
16 50
C = {10, 20 …… 50}
P(C) =
5 50
A B = {6, 12, 18, …… 48}
P(A B) =
8 50
B C = {30}
P(B C) =
1 50
A C = {10, 20, …… 50}
P(A C) =
5 50
And A B C = {30}
P(A B C) =
1 50
Required probability = P(A B C) = P(A) + P(B) + P(C) – P(A B) – P(B C) – P(A C) + P(A B C) =
25 16 5 8 1 5 1 33 50 50 50 50 50 50 50 50
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(School/Board Exams.) Solutions 8
40. (i)
C2
20
8
(ii)
C2
6
6
(ii)
C3
4
C4
15
6
4 19
3 10
1 6
C3
10
(ii)
C2 6C1 4C3
9
42. (i)
C1 4C2 10
(iii)
C4
C3 42 6 455 65
C2 9C2 36 1365 91
(iii)
1 65
(iv)
48 91
43. (i)
C2
C3
10
109
14 95
C1 5C1 20
41. (i)
Probability
P(At least one of E1 and E2) = P(E1 E2) = P(E1) + P(E2) – P(E1 E2) = 0.50 + 0.40 – 0.15 = 0.75
(ii)
P(neither E1 nor E2 occurs) = P(E1 E2) = P(E1 E2) = 1 – P(E1 E2) = 1 – 0.75 = 0.25
30
44.
C3 25 C3 30
= 45. (i)
C3
88 203 A = getting an odd number on the first die = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6), (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)}
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Probability
(ii)
(School/Board Exams.) Solutions
A or B = Getting any number on the first die S {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)} (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
(iii) B and C = {(1, 1), (1, 2), (1, 3), (1, 4) (3, 1), (3, 2)} (iv)
A and D = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (6, 1), (6, 2), (6, 3), (6, 4)}
SECTION - B NCERT Questions Exercise-16.1 1.
When a coin is tossed once, there are 2 possible outcomes, ‘Head’ and ‘Tail’, usually represented by H and T. When a coin is tossed 3 times, then there are 23 = 8 possible outcomes. If S is the sample space of this experiment, then S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Number of times a coin is tossed 1
2 3 HHH 2.
Outcomes
H
HH
T
HT
HHT HTH
TH HTT THH
THT TTH
TT TTT
A die has six faces, each face bearing one of the numbers from {1, 2, 3, 4, 5, 6}. So, when a die is rolled once, then there are six possible outcomes. When the die is rolled twice, there are 6 × 6 = 36 possible outcomes. If S is the sample space of this experiment, then S = {(x, y): x, y {1, 2, 3, 4, 5, 6}}
i .e.,
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3),(3, 4),(3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5),(5, 6), (6, 1), (6, 2),(6, 3), (6, 4), (6, 5), (6, 6)}
3.
When a coin is tossed once, there are 2 possible outcomes ‘Head’ and ‘Tail’ usually written as H and T. When a coin is tossed four times, there are 2 2 2 2 = 24 = 16 possible outcomes. If S is the sample space of this experiment, then {HHHH, HHHT, HHTH, HHTT, HTHH,HTHT, S HTTH,HTTT, THHH, THHT, THTH, THTT, TTHH,TTHT, TTTH, TTTT}
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(School/Board Exams.) Solutions
4.
Probability
111
When a coin is tossed, there are two possible outcomes H and T and when a die is rolled once, then there are six possible outcomes 1, 2, 3, 4, 5 and 6.
When ‘a coin is tossed and a die is thrown’, then the sample space contains 2 6 = 12 outcomes.
If S is the sample space, then
S
{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6) (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
H
5.
T
1 2 3 4 5 6 1 2 3 4 5 6 When a coin is tossed, it will show either a ‘head’ or a ‘tail’. When the coin shows up a ‘head’, a die is rolled which may show up any one of the six numbers 1, 2, 3, 4, 5, 6. H
1 2 3 4 5 6 Let S denotes the sample space of the experiment, then S = {T, (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)} 6.
Let the boys and girls in room X be denoted by B1, B2; G1, G2 and those in room Y be denoted by b1, g1, g2, g3. A room out of the two can be selected in two ways, either the room X is selected or the room Y is selected. Corresponding to either case, there are four ways of selecting a person.
X
B1 B2
Y
b1
G1 G2
g2
g1
g3
If S is the sample space, then S
7.
{( X , B1 ), ( X , B2 ), ( X , G1 ), ( X , G2 ), (Y , b1 ), (Y , g1 ), (Y , g 2 ), (Y , g3 )}
There are 3 ways of picking up a die from the bag. Any one of red, white or blue die may be selected. Corresponding to each way of doing so there are six possible outcomes 1, 2, 3, 4, 5, 6. Let the three dice be represented by R, W and B respectively.
R
1
2 3
W
4 5 6
1
2 3
B
4 5 6
1
2 3
4 5 6
If S is the sample space, then {(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6), (W , 1), S (W , 2), (W , 3), (W , 4), (W , 5), (W , 6), (B, 1), (B, 2), (B, 3), (B, 4), (B, 5), (B, 6)} Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
112 8.
Probability
(i)
(School/Board Exams.) Solutions
Here, there are four possibilities (1) Both the children are boys (2) First child is a boy and second child is a girl (3) First child is a girl and second child is a boy (4) Both the children are girls In this case, the sample space is S = {BB, BG, GB, GG}
(ii)
In this case, there are 3 possible cases: (1) There is no girl child. (2) There is only one girl child (3) Both the children are girls.
So, in this case, the sample space is S = {0, 1, 2}. 9.
S = {(R, W1), (R, W2), (R, W3), (W1, R), (W1, W2), (W1, W3), (W2, R), (W2, W1), (W2, W3), (W3, R), (W3, W1), (W3, W2)}
10. It is a two stage experiment. First stage results in either ‘head’ or ‘tail’. When the coin shows up head, then it is tossed again showing up either a head or a tail and if the first toss shows up a tail, then a die is rolled once which may show up any one of the six numbers 1, 2, 3, 4, 5, 6
H
T
HH
HT
T1 T2 T3 T4 T5 T6
If S is the sample space, then S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}. 11. This experiment is analogous to the experiment of “tossing a coin thrice” or tossing “three coins once”. The only difference is that here we use symbols D and N in place of H and T. Here also, there are 23 = 8 possible outcomes.
D
DD DDD
N
DN
DDN DND
ND
DNN NDD
NN
NDN NND
NNN
Here, the sample space is S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN} Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
(School/Board Exams.) Solutions
Probability
113
12. It is a three stage experiment
First stage
Second stage 1
Third stage
T
H
1
2
2 3
3
4
4 5 6
1 2 3
5
6
4 5 6 1 2 3
4 5 6
{T, (H, 1), (H, 3), (H, 5), (H, 2, 1), (H, 2, 2), (H, 2, 3), (H, 2, 4), (H, 2, 5), (H, 2, 6 ), (H, 4, 1), (H, 4, 2), (H, 4, 3), Here the sample space is S (H, 4, 4), (H, 4, 5), (H, 4, 6), (H, 6, 1), (H, 6, 2), (H, 6, 3), (H, 6, 4), (H, 6, 5), (H, 6, 6 )} Note that the sample space contains 1 + 3 + 18 = 22 outcomes. 13. Here, the first slip has four possibilities and second only three. So, there are 4 3 = 12 possible outcomes. The second slip must show a number different from one on the first slip.
First slip
Second slip
1
2
2
3
4
1
3
3
4
1
2
4
4
1
2
3
If S is the sample space, then S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}. 14. It is a two stage experiment. When a die is rolled, it may show up any one of this six numbers 1, 2, 3, 4, 5, 6.
First stage 1
2
3
4
5
6
Second stage HH HT TH TT H T HH HT TH TT H T HH HT TH TT H T
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Probability
(School/Board Exams.) Solutions
When, the die shows up an odd number, a coin is tossed twice and hence, corresponding to each of 1, 3 and 5 there are four possibilities. When the die shows up an even number, a coin is tossed once and hence, corresponding to each of 2, 4 and 6 there are only two possible outcomes. In this case, the sample space is
{2H, 2T, 4H, 4T, 6H, 6T, S 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT} Note that there are 18 possible outcomes. 15. Let the balls in the box be represented by R1, R2 and B1, B2, B3. It is a two stage experiment.
H
Stage I
Stage II
1 2 3
4
T
5 6
R1
R2 B1
B2
B3
Sample space S is given by S
{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6) (T, R1 ), (T, R 2 ), (T, B1 ), (T, B2 ), (T, B3 )}
16. If the die shows up 6, we stop. If it shows any other number, then the die is tossed again. If the second toss shows up 6, we stop. If the second toss shows any other number, then we toss the die again. In this experiment dice may be tossed once, twice, thrice, four times, …… any number of times. Here, the sample space is
{6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), S (1, 1, 6), (1, 2, 6),(1, 3, 6), (1, 4, 6), (1, 5, 6), (2, 1, 6), (2, 2, 6),} Exercise-16.2 1.
Here, the sample space is S = {1, 2, 3, 4, 5, 6}. The event E : ‘die shows 4’ and F : ‘die shows an even number’ i.e., E = {4} and F = {2, 4, 6} As
2.
E F = {4} , therefore, E and F are not mutually exclusive.
When a die is rolled once, the sample space is S = {1, 2, 3, 4, 5, 6}. (i)
A : ‘a number less than 7’,
A = {1, 2, 3, 4, 5, 6} = S ; A sure event
(ii)
B : ‘a number greater than 7’,
B = { }, an impossible event
Note that there is no outcome in the sample space which is greater than 7. (iii) C : ‘a multiple of 3’,
C = {3, 6}.
(iv)
D : ‘a number less than 4’,
D = {1, 2, 3}.
(v)
E : ‘an even number greater than 4’,
E = {6}.
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(vi)
F : ‘a number not less than 3’,
Probability
i.e.,
115
F : ‘a number greater than or equal to 3
F = {3, 4, 5, 6} A B = S { } = S = {1, 2, 3, 4, 5, 6}, A B = S { } = { } B C = { } {3, 6} = {3, 6}, E F = {6} {3, 4, 5, 6} = {6}, D E = {1, 2, 3} {6} = { }, A – C = S – {3, 6} = {1, 2, 3, 4, 5, 6} – {3, 6} = {1, 2, 4, 5} D – E = {1, 2, 3} – {6} = {1, 2, 3} E F= {6} {3, 4, 5, 6} = {6} {1, 2} = { } and F= {3, 4, 5, 6} = {1, 2}. 3.
When a pair of dice is rolled once, then the sample space contains 6 6 = 36, ordered pairs of the type (x, y), where x is the number that comes up on the first die and y is the number that comes up on the second die. So, the sample space is S = {(x, y) : x, y {1, 2, 3, 4, 5, 6}}. (i)
A : ‘the sum is greater than 8’ i.e., A : ‘the sum is 9, 10, 11 or 12’
(ii)
A = {(3, 6), (6, 3), (4, 5), (5, 4), (6, 4), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)} B : ‘2 occurs on either die’
(iii)
B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)} C : ‘the sum is atleast 7 and a multiple of 3’
i.e., C : ‘the sum is 9 or 12’
C = {(3, 6), (6, 3), (4, 5), (5, 4), (6, 6)}
Here, A B = , B C = , therefore, A and B are mutually exclusive and also B and C are mutually exclusive. Note that A and C are not mutually exclusive as A C 4.
Here, the sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} When three coins are tossed once or one coin is tossed thrice, the sample space is same Here, A : ‘getting three heads’ i.e., A = {HHH} B : ‘getting two heads and one tail’ i.e., B = {HTH, HHT, THH} C : ‘getting three tails’ i.e., C = {TTT} and D : ‘getting head on the first coin’ i.e., D = {HHH, HHT, HTH, HTT} (i)
Here,
A B = A C = B C = D C =
Hence,
A and B are mutually exclusive, A and C are mutually exclusive, B and C are mutually exclusive,
and
C and D are mutually exclusive.
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Probability
(ii)
(School/Board Exams.) Solutions
As A and C contain only one outcome, therefore, A, C are simple events.
(iii) Both B and D are compound events as either of them contains more than one outcomes. 5.
In this case also, the sample space of the experiment is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} When a single coin is tossed thrice or three coins are tossed once, the sample space is the same. (i)
Consider the events E1 : ‘one head and two tails show up’ and E2 : ‘one tail and two heads show up’ i.e., E1 = {HTT, THT, TTH} and E2 = {HTH, THH, HHT} Here, E1 and E2 are mutually exclusive
(ii)
Consider the events E1 : ‘not more than one head shows up’ E2 : ‘exactly two heads show up’ and E3 : ‘three heads show up’ i.e., E1 = {TTT, HTT, THT, TTH} E2 = {HHT, THH, HTH} E3 = {HHH} In this case, E1 E2 = E2 E3 = E3 E1 = and E1 E2 E3 = S.
The three events E1, E2 and E3 are mutually exclusive and exhaustive
(iii) Consider the events E1 : ‘not more than one head shows up’ and E2 : ‘no head show up’ i.e., E1 = {TTT, HTT, THT, TTH} and E2 = {TTT} Here, E1 and E2 are not mutually exclusive. (iv)
Events E1 and E2 of part (i) are mutually exclusive but not exhaustive as E1 E2 = but E1 E2 S
(v)
Consider the events E1 : ‘only one head shows up’ E2 : ‘exactly two heads show up’ and E3 : ‘three heads show up’ i.e., E1 = {HTT, THT, TTH} E2 = {THH, HHT, HTH} and E3 = {HHH}, then E1 E2 = E2 E3 = E3 E1 = but E1 E2 E3 S
The three events E1, E2, E3, are mutually exclusive but not exhaustive.
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(School/Board Exams.) Solutions
6.
Probability
117
Here, the sample space contains 6 6 = 36 simple events Sample space is S = {(x, y) : x, y {1, 2, 3, 4, 5, 6} }
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), i.e., S
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Given
A : “getting an even number on first die” B : “getting an odd number on first die”
and
C : ‘getting the sum of number on the die less than equal to 5
i.e.,
{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), A (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), B (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
{(1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1),
and
C
(i)
A = { : A}
(1, 4), (2, 3), (3, 2), (4, 1)}
(sum is less than or equal to sum is 2, 3 or 4)
= S – A = B (note that A B = and A B = S) (ii)
not B = B = { : B} =S–B=A
(iii) A or B = A B = S (iv)
A and B = {: A and B} ={}
(v)
A but not C = A – C = {: A and C} =
(vi)
{(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B or C = B C = {: B or C}
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), = (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (2, 1), (2, 2), (2, 3), (4, 1)} (vii) B and C = B C = { : B and C} = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)} Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Probability
(School/Board Exams.) Solutions
(viii) A BC = (A B)C = {A A}C = A C = A but not C = { : A and C} =
{(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Alternatively, A B C = (A B)C = (A B) –C = (A – B) –C = A – C (∵ A and B have no element in common, therefore, A – B = A) 7.
(i)
True (∵ A B = )
(ii)
True (∵ A B = and A B = S)
(iii) True (∵ A = S – B B) (iv)
False (∵ A C = {(2, 1), (2, 3), (4, 1)} )
(v)
False (∵ A B = A A = A )
(vi)
False (∵ A C = B C and also B C = A C
Exercise-16.3 1.
(a)
Here, each assigned value is non-negative and the sum of all the values = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.00 Therefore, given assignment of probabilities is valid.
(b)
Here, also each assigned value is non-negative and sum of all the values =
1 1 1 1 1 1 1 1 7 7 7 7 7 7 7
Therefore, given assignment of probabilities is valid (c)
In this case, the sum of assigned values = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 1 Therefore, the given assignment of probabilities is not valid.
(d)
In this case, values assigned to 1 and 5 are negative, therefore, the given assignment of probabilities is not valid.
(e)
In this case, the sum of assigned values =
1 2 3 4 5 6 15 36 1 14 14 14 14 14 14 14 14
Therefore, the given assignment of probabilities is not valid. Alternatively, the value assigned to 7 is 2.
15 > 1, therefore, the given assignment of probabilities is not valid. 14
When a coin is tossed twice, the sample space S contains four equally likely outcomes HH, HT, TH and TT, i.e., S = {HH, HT, TH, TT} Let E : ‘atleast one tail occurs’, then E = {HT, TH, TT}
P (E )
Number of outcomes favourable to E 3 Total number of outcomes 4
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3.
Probability
119
When a dice is thrown, there are six possible outcomes 1, 2, 3, 4, 5, 6 Sample space S = {1, 2, 3, 4, 5, 6} All the outcomes are equally likely. (i)
Let E1 : ‘A prime number will appear, then E1 = {2, 3, 5} Hence, P (E1 )
(ii)
Number of outcomes favourable to E1 3 1 Total number of outcomes 6 2
Let E2 : ‘A number greater than or equal to 3 will appear’, then E2 = {3, 4, 5, 6} Hence, P (E2 )
Number of outcomes favourable to E2 4 2 Total number of outcomes 6 3
(iii) Let E3 : ‘A number less than or equal to one will appear’, then E3 = {1} Hence, P (E3 ) (iv)
Number of outcomes favourable to E3 1 Total number of outcomes in S 6
Let E4 : ‘A number more than 6 will appear’, then E4 = { }, i.e., E4 is an impossible event
(∵ There is no outcome greater than 6)
Hence, P(E4) = 0 (v)
Let E5 : ‘A number less than 6, will appear, then E5 = {1, 2, 3, 4, 5} Hence, P(E5) =
4.
Number of outcomes favourable to E5 5 Total number of outcomes in S 6
When a card is taken out from a deck of 52 cards, there are 52 equally likely outcomes. (a)
Sample space contains 52 sample points
(b)
P (card drawn is an ace of spades) =
(c)
(i)
1 52
(∵ There is only one card which is an ace of spades)
P(Card drawn is an ace) =
4 1 52 13
(∵ There are four aces in the deck)
(ii) P(card drawn is a black card) = 5.
26 1 52 2
(∵ There are 26 black cards, 13 of spades and 13 of clubs)
Here, the sample space is
S
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
1
1 2 3
6
4
5 6
1 2 3
4
5 6
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Probability
(School/Board Exams.) Solutions
Thus, the sample space contains 2 6 = 12 equally likely outcomes (i)
P(sum of the numbers is 3) = P({1, 2}) =
1 . 12
(∵ Only one outcome (1, 2) is favourable to the event in reference) (ii)
P(sum of the number is 12) = P({(6, 6)}) =
1 . 12
(∵ Only one outcome (6, 6) is favourable to the event in reference) 6.
There are total 10 members of which only six are women
7.
P(a woman member is selected) =
6 3 . 10 5
When a coin is tossed four times, there are 24 = 16 possible equally likely outcomes. The sample space in this case is {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, S HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
Let E() be the amount of money corresponding to an outcome S, then E(HHHH) = Rs(1 + 1 + 1 + 1) = Rs 4 E(HHHT) = Rs(1 + 1 + 1 – 1.50) = Rs 1.50 E(HHTH) = Rs(1 + 1 – 1.50 + 1) = Rs 1.50 E(HHTT) = Rs(1 + 1 – 1.50 – 1.50) = – Rs 1 E(HTHH) = Rs(1 – 1.50 + 1 + 1) = Rs 1.50 E(HTHT) = Rs(1 – 1.50 + 1 – 1.50) = – Rs 1 E(HTTH) = Rs(1 – 1.50 – 1.50 + 1) = – Rs 1 E(HTTT) = Rs(1 – 1.50 – 1.50 – 1.50) = – Rs 3.50 E(THHH) = Rs(– 1.50 + 1 + 1 + 1) = Rs 1.50 E(THHT) = Rs(– 1.50 + 1 + 1 – 1.50) = – Rs 1 E(THTH) = Rs(– 1.50 + 1 – 1.50 H) = Rs –1 E(THTT) = Rs(– 1.50 + 1 – 1.50 – 1.50) = – Rs 3.50 E(TTHH) = Rs(– 1.50 – 1.50 + 1 + 1) = – Rs 1 E(TTHT) = Rs(– 1.50 – 1.50 – 1.50) = – Rs 3.50 E(TTHT) = Rs(– 1.50 – 1.50 – 1.50 + 1) = – Rs 3.50 E(TTTT) = Rs(– 1.50 – 1.50 – 1.50 – 1.50) = – Rs 6 Thus, we see that there are five possible amounts of money, gain of Rs. 4, gain of Rs. 1.50, loss of Rs 1, loss of Rs 3.50 and loss of Rs 6. Case-I
Four heads appear P(HHHH) =
1 1 1 1 1 2 2 2 2 16
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(School/Board Exams.) Solutions
Case-II
Probability
121
3 heads, 1 tails Required probability =
4
C3
1 1 1 1 1 = 2 2 2 2 4
Total money = 3 1 – 1 1.5 = Rs 1.5 Case-III 2 heads, 2 tails 4
Probability =
3 1 1 1 1 = 8 2 2 2 2
C2
Total money = 2 1 – 2 1.5 =–1 Case-IV 1 heads, 3 tail 4
Probability =
C1
1 1 1 1 2 2 2 2
1 4
=
Total money = 1 1 – 3 1.5 = – 3.5
Probabilities of various amounts of money are as follows:
Amount
Rs 4 Rs 1.50 –Rs 1 –Rs 3.50 –Rs 6 1 1 3 1 1 Probability 16 4 8 4 16 Case-V 4 tails P (T T T T)
1 16
Total money = 4 –1.5 = – 6 8.
When three coins are tossed, there are 23 = 8 possible equally likely outcomes. The sample space in this case is
S
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 1 8
(i)
P(3 heads) = P{HHH} =
(ii)
P(2 heads) = P({HHT, HTH, THH}) =
3 8
(iii) P(at least 2 heads) = P(either 2 heads or 3 heads) = P({HHT, HTH, THH, HHH}) =
4 1 8 2
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Probability
(iv)
(School/Board Exams.) Solutions
P(at most 2 heads) = P(either no head or 1 head or 2 heads) = P({TTT, HTT, THT, TTH, HHT, HTH, THH}) =
8
Alternatively, P(at most 2 heads) = 1 – P(3 heads) = 1 – P(HHH) = 1 (v)
1 7 8 8
1 8
P(no head) = P({TTT}) =
1 8 (vii) P(exactly two tails) = P({THT, TTH, HTT})
(vi)
P(3 tails) = P(TTT) =
=
3 8 1 8
(viii) P(no tail) = P({HHH}) =
(ix) P(at most two tails) = 1 – P(3 tails) = 1 – P({TTT}) = 1 9.
Given P(A) =
1 7 8 8
2 11
P(not A) = 1 – P(A) = 1 –
2 9 11 11
10. The word ASSASSINATION consists of 13 letters, 6 vowels A, A, A, I, I, O and 7 consonants S, S, S, S, N, T, N 6 13
(i)
P(a vowel is chosen) =
(ii)
P(a consonant is chosen) =
7 13
11. Out of 20, a person can choose 6 natural numbers in 20C6 ways. Out of these there is only one choice which will match the six numbers already fixed by the committee
P(the person wins the prize) 1
6 20 6 20
=
20
=
6 14 6 5 4 3 2 1 14 20 20 19 18 17 16 15 14
=
2 1 19 17 16 15 19 17 8 15
=
1 38760
C6
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12. (i)
Probability
123
Since, A B A, therefore we must have P(A B) P(A). But, we are given that P(A) = 0.5 and P(A B) = 0.6 which is not consistent.
(ii)
Here, P(A) + P(B) = 0.5 + 0.4 = 0.9 and P(A B) = 0.8 < P(A) + P(B), therefore, these probabilities are consistently defined (∵ P(A B) = P(A) + P(B) – P(A B) < P(A) + P(B))
13. (i)
P(A B) = P(A) + P(B) – P(A B) =
(ii)
1 1 1 5 3 1 7 3 5 15 15 15
P(A B) = P(A) + P(B) – P(A B)
P(B) = P(A B) + P(A B) – P(A) = 0.6 + 0.25 – 0.35 = 0.5 P(A B) = P(A) + P(B) – P(A B)
(iii)
P(A B) = P(A) + P(B) – P(A B) = 0.5 + 0.35 – 0.7 = 0.15
Complete table is
P(A)
P(B)
P(A B)
1 5
1 15
7 15
0.25
0.6
(i)
1 3
(ii)
0.35
0.5
(iii)
0.5
0.35
0.15
P(A B)
0.7
14. A and B are mutually exclusive events, therefore, A B =
P(A B) = 0
P(A B) = P(A) + P(B) – P(A B) =
15. (i)
(ii)
P(A or B) =
3 1 4 0 5 5 5
4 5
P(E or F) = P(E F) = P(E) + P(F) – P(E F) =
1 1 P (E and F ) 4 2
=
1 1 1 2 4 1 5 4 2 8 8 8
P(not E and not F) = P(E F) = P(E F)
(De Morgan’s laws)
= 1 – P(E F) = 1
5 3 = 8 8
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Probability
(School/Board Exams.) Solutions
16. P(not E or not F) = 0.25
P(E F) = 0.25
P((E F)) = 0.25
1 – P(E F) = 0.25
P(E F) = 1 – 0.25 = 0.75 0
E and F are not mutually exclusive
17. (i)
P(not A) = 1 – P(A) = 1 – 0.42 = 0.58
(ii)
P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A B) = P(A) + P(B) – P(A B) = P(A) + P(B) – P(A and B) = 0.42 + 0.48 – 0.16 = 0.74 18. P(student is studying mathematics or Biology) = P(student is studying Mathematics) + P(student is studying Biology) – P(student is studying both Mathematics and Biology) (∵ P(A B) = P(A) + P(B) – P(A B)) =
40 30 10 60 0.6 100 100 100 100
19. Let E1 : student passes first examination E2 : student passes second examination We are given P(E1) = 0.8, P(E2) = 0.7 and P(at least one of E1 and E2) = 0.95
P(E1 or E2) = 0.95
P(E1) + P(E2) – P(E1 and E2) 0.95
0.8 + 0.7 – P(E1 and E2) = 0.95
P(E1 and E2) = 0.8 + 0.7 – 0.95 = 0.55
P(students passes both examination = 0.55) E1 : student passes in English
20.
and E2 : student passes in Hindi We are given that P(E1 and E2) = 0.5, i.e., P(E1 E2) = 0.5 P(neither E1 nor E2) = 0.1
P(E1 and E2) = 0.1
P(E1 E2) = 0.1
P(E1 E2) = 0.1
1 – P(E1 E2) = 0.1
1 – {P(E1) + P(E2) – P(E1 E2)} = 0.1
1 – {0.75 + P(E2) – 0.5} = 0.1
1 – 0.75 – P(E2) + 0.5 = 0.1
P(E2) = 0.75 – 0.1 = 0.65.
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Probability
125
21. Let E1 : student opted for NCC E2 : student opted for NSS, then P(E1) =
P (E 2 )
30 1 , 60 2
32 8 and 60 15
P (E1 and E2 )
(i)
P(E1 E2) = P(E1) + P(E2) – P(E1 and E2) =
(ii)
24 2 60 5
1 8 2 15 16 12 19 2 15 5 30 30
P(student opted NCC or NSS) =
19 30
P(student opted neither NCC nor NSS) = P(E1 and E2) = P(E1 E2) = P((E1 E2)) = 1 – P(E1 E2) = 1
=
19 30
(Using part (i))
11 30
(iii) P(student has opted NSS but not NCC) = P(E2 but not E1) = P(E2 E1) = P(E2 – E1) = P(E2 – E2 E1) = P(E2) – P(E2 E1) =
8 2 15 5
=
86 2 15 15
{∵ E2 E1 E2}
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Probability
(School/Board Exams.) Solutions
Miscellaneous Exercise 1.
5 marbles out of 10 + 20 + 30 = 60 can be drawn in (i)
5 blue marbles out of 20 can be drawn in 20
(ii)
P(all 5 marbles are blue) =
60
20C 5
60C 5
ways.
ways
C5 C5
=
20 19 18 17 16 5 5 60 59 58 57 56
=
1 1 18 17 16 2 17 34 = 3 3 59 58 56 59 29 7 11977
P(atleast one green marble is chosen) = 1 – P(no green marble is chosen) = 1
60
C5
(5 Non-green marbles can be selected out of 10 + 20 = 30 in
C5
= 1
30 29 28 27 26 1 1 1 27 26 1 60 59 58 57 56 2 2 2 59 57
= 1
9 13 117 1 4 59 19 4484
4 cards can be drawn of 52 in
52C 4
ways
3 spades (out of 13) and one ace (out of 13) can be selected in
=
=
13
C3 52
C1
C4
13C 1
ways
13 12 11 13 3 1 52 51 50 49 4
13 2 11 13 22 13 286 52 51 50 49 17 25 49 20825 1 2 3 4
Here, P(1) =
(i)
13C 3
P(3 diamond and one spade is drawn) 13
3.
30C ) 5
4484 117 4367 4484 4484
= 2.
30
2 1 6 3
(∵ Two faces out of six are marked 1)
P (2)
3 1 and 6 2
(∵ Three faces out of six are marked 2)
P (3)
1 6
(∵ Only one face out of six bears the number 3)
P (2)
1 2
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(School/Board Exams.) Solutions
(ii)
Probability
P(1 or 3) = P(1) + P(3) =
1 1 3 1 3 6 6 2
(iii) P (not 3) = 1 – P(3) = 1 – 4.
(a)
1 5 6 6
One ticket out of 10,000 can be selected in
10000C 1
One ticket with prize can be selected out of 10 in
= 10000 ways
10C 1
= 10 ways
10 1 10000 1000
P(getting a prize) =
P(not getting a prize) = 1 – P(getting a prize) = 1
(b)
1 999 1000 1000
Two tickets out of 10,000 can be selected in
10000C 2
ways.
Two tickets (not winning a prize) out of 10,000 – 10 = 9990 can be selected in
9990
P(not getting a prize) =
= (c)
9990 9989 1108779 10000 9999 1111000
9990
P(not getting a prize) =
10000C 10
ways. 9990C 10
ways
C10
10000
C10
Two sections of 40 and 60 can be formed out 100 in (a)
ways.
C2
10 tickets out 10000 can be selected in
9990C 2
C2
10000
10 tickets (not winning a prize) out of 10000 – 10 = 9990 can be selected in
5.
127
100C 60
(or
100C ) 40
ways.
Both you and your friend can go either to one section containing 40 students or to the other section containing 60 students, which can happen in 98C38 + 98C40 ways (When both go to the section of 40, we have to select 38 out of 98 which can be done in 98C38 ways and when they go to the other section, we have to select 40 out of 98 students which can be done in 98C 40 ways). 98
P(both enter the same section) =
=
C38 100
98
C40
C40
98 98 38 60 40 58 100 60 40
60 40 98 ⎧ 1 1 ⎫ ⎨ ⎬ 100 ⎩ 38 60 40 58 ⎭
98 ⎧ 40 60 ⎫ = 100 ⎨ 38 58 ⎬ ⎩ ⎭ =
40 39 60 59 1560 3540 5100 17 100 99 9900 9900 33
Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
128
Probability
(b)
(School/Board Exams.) Solutions
When you and your friend enter into different sections, then it can happen in
98C 39
2C1 ways.
(∵ For the section containing 40 students, we have to select 39 out of remaining 98 and one out of you and your friend) 98
Required probability =
C39 2C1 100
C40
2
98
100
C39
(∵ 2C1 = 2)
C40
2 98 60 40 2 40 60 48 16 = 59 39 100 100 99 99 33
6.
Let the three letters be denoted by L1, L2, L3, and three envelopes by E1, E2, E3. Total number of ways of putting the letters into three envelopes is 3P3 = 3 6 The number of ways in which none of the letter is put into proper envelope is 2.
E1
E2
E3
Required probability
L2
L3
L1
= P(atleast one of the letters is put into right envelope)
L3
L1
L2
= 1 – P(none of the letters is put into right envelope) = 1 7.
(i)
2 1 2 1 6 3 3
P(A B) = P(A) + P(B) – P(A B) = 0.54 + 0.69 – 0.35 = 1.23 – 0.35 = 0.88
(ii)
P(AB) = P((A B)) = 1 – P(A B) = 1 – 0.88
(Using part (i))
= 0.12 (iii) P(A B) = P(A – B) = P(A – A B ) = P(A) – P(A B)
(∵ A B A)
= 0.54 – 0.35 = 0.19 (iv)
P(BA) = P(B – A) = P(B – A B) = P(B) – P(A B) = 0.69 – 0.35 = 0.34
8.
Let E1 : Spokes person is a male and E2 spokes person is over 35 years, then
P (E1 )
3 2 and P (E2 ) 5 5
Also P(E1 E2) = P(selecting a male over 35 years) =
1 5
P(selecting a male person or a person over 35 years) = P(E1 or E2)
Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
(School/Board Exams.) Solutions
Probability
129
= P(E1 E2) = P(E1) + P(E2) – P(E1 E2) = 9.
(i)
3 2 1 4 5 5 5 5
When repetition of digits is allowed Th
H
T
U
2 ways 5 ways 5 ways 5 ways
Left most place can be filled in two ways as only 5 or 7 can be placed there ( Number to be formed are greater than 5000) Each of the remaining three places can be filled in 5 ways.
Number of numbers that can be formed = 2 5 5 5 = 250 – 1 = 249 (excluding 5000) If the number is divisible by 5, then units place is to be filled Th H T U 2 ways 5 ways 5 ways 2 ways
With either 0 or 5. Also, the thousand’s place can be filled in two ways as only 5 and 7 are to be placed there. Each of the remaining two places can be filled in 5 ways.
Number of numbers that are divisible by 5 = 2 5 5 2 = (100 – 1) = 99 (excluding 5000) Required probability =
(ii)
99 249
When the repetition of digits is not allowed: As the number to be formed is more than 5000, leftmost place (i.e., thousand’s place) can be filled in two ways only as 5 or 7 can be placed there. The remaining three places can be filled in with remaining four digits in 4P3 ways.
Number of numbers (> 5000) that can be formed = 2 4P3 = 2 4 3 2 = 48 The numbers which are divisible by 5 have either 0 or 5 in the units place Case-I In this case 5 is fixed in units place, 7 is fixed in thousands place and the remaining two places can be filled in with remaining 3 digits in 3P2 = 3 2 = 6 ways
Number of numbers of this kind = 6 Case-II In this case 0 is fixed in unit’s place. Thousand’s place can be filled in two ways as either 5 or 7 can be placed there. The remaining two places can be filled in with remaining three digits in 3P2 = 3 2 = 6 ways. Number of numbers of this kind = 2 6 = 12
Number of numbers divisible by 5 = 6 + 12 = 18
Required probability =
18 3 . 48 8
Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
130
Probability
(School/Board Exams.) Solutions
10. The number of all possible sequences = 10 10 10 10 = 10000 (∵ Each of the four wheel has 10 digits marked on it) The lock is to be opened with a sequence of 4 distinct digits. 10P 4
Number of such sequences =
= 10 9 8 7 = 5040 (10 digits can be arranged in a line 4 at a time in
Required probability =
10P 4
ways)
1 5040
SECTION - C Model Test Paper Very Short Answer Type Questions : 1.
S = {(T, R1), (T, R2), (T, B1), (T, B2), (T, B3) (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
2.
It is a sure event
3.
The prime numbers between 1 and 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
4.
Probability of getting atleast 52 wednesdays in a leap year is 1
Required probability =
15 3 50 10
Total number of cases = 6 6 E : Event that the sum is prime {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), Favourable events to E = (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
P (E )
15 5 36 12
5.
Required number of elements = 7 × 6 = 42
6.
Total number of outcomes = 6 6 6 = 216 Favourable outcomes = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)}
Required probability =
6 1 216 36
Short Answer Type Questions : 7.
Each of the five places can be filled in 7 ways.
Total number of numbers that can be formed = 7 7 7 7 7 = 75
Since, None of the digits is repeated
Number of possible numbers that can be formed is 7P5
Required probability =
7
P5
7
5
360 2401
Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
(School/Board Exams.) Solutions
8.
Probability
Total number of ways of arranging the letters of the word MUMMY is
131
5! and when the letters at extreme ends 3!
are M’s, then the number of ways is 3! [M……M] 6
9.
Required probability =
C2 8C3 14
C5
3! 3! 3 5! 10
60 143
10. A Event “number on the card is divisible by 3” B Event “number on the card is divisible by 7” P(A B) = P(A) + P(B) – P(A B)
P(A B) =
5 2 0 17 17
P(A B) =
7 17
11. Number of elements in sample space is
10C 2
= 45
Selected two balls will be of different colours if either one is white and one red or one is red and one green or one is green and one white. Hence number of different ways selecting two balls of different colours = 2C1 × 3C1 + 3C1 × 5C1 + 5C1 × 2C1 = 31
Required probability =
31 45
12. A and B are mutually exclusive
P(A B) = 0
Now, P(A B) = P(A) + P(B) = [1 P ( A)] P (B ) 0.75 = [1 – 0.6] + P(B) 13.
P(B) = 0.35
5! 4! 1 8! 14
14. Since, A and B mutually exclusive and exhaustive events,
P(S) = P(A B) = P(A) + P(B)
1 = P(A) +
1
P ( A)
3 P ( A) 2
[∵ 3P(A) = 2P(B)]
5 P ( A) 2 2 5
Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
132
Probability
(School/Board Exams.) Solutions
15. A Event “number is divisible by 6” B Event “number is divisible by 8”
P(A) =
33 25 , P(B) = , 200 200
P(A B) =
8 200
A B Numbers which are divisible by both 6 and 8 i.e., by 24 Now, P(A B) = P(A) + P(B) – P(A B) =
33 25 8 200 200 200
=
50 1 200 4
Long Answer Type Questions : 16. Total number of different ways selecting three children =
12C 3
= 220
Number of different ways so that selection of three children consists at least 2 girls = (number of ways selecting two girls and one boy) + (number of ways selecting three girls) = 6C2 × 6C1 + 6C3 = 110 17. (i)
Required probability =
110 1 220 2
P(not A grade) = 1 – P(A grade) = 1 – 0.4 = 0.6
(ii)
P(B or C grade) = P(grade B) + P(grade C) = 0.35 + 0.15 = 0.5
(iii) P(atmost C grade) = P(grade C) + P(grade D) = 0.15 + 0.10 = 0.25 18. (i)
1 25
(ii)
2 25
(iii)
4 25 20
19. (i)
(iii)
100
1
80
C10 C10 80
C10
100
C10
(ii)
80
(iv)
C10
100
C10
C10
100
C10
Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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