# Probability and Statistics

August 8, 2017 | Author: AkashGaurav | Category: Mode (Statistics), Covariance, Variance, Median, Arithmetic Mean

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#### Description

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PROBABILITY AND STATISTICS INTRODUCTI ON

Probability of occurrence of A is given by

Probability is the measure of degree of certainity or uncertainity of occurrence of an event. The set of all possible outcomes is called sample space of that experiment and is denoted by S. Example : Sample space of tossing a coin : S = {H, T} H-head, T-tail. Sample space of tossing 2 coins : S = {(H, H), (H, T), (T, H), (T, T)} Sample space of throwing a die : S {1, 2, 3, 4, 5, 6} Probability of an event is denoted by P(E). 0 £ P(E) £ 1 Probability of sample space, P(S) = 1.

æAö ÷ P(Bi ) i=1 è i ø P(A/Bi)-Probability of occurrence of A when Bi has already happened.

Notations Used in Probability · Probability of happening of events A or B : P(A È B) or P(A + B). · Probability of happening of events A and B : P(A Ç B) or P(AB). Mutually exclusive events Event whose occurrences is independent of occurrence of another events is said to be mutually exclusive. They don’t occur simultaneously i.e. A Ç B = f E.g. : tossing a coin, throwing a die. Equally likely events If an event cannot happen in performance to other such events is said to be equally likely. Independent events Two events are said to be independent if happening or failure of one does not effect the happening or failure of the others. Exhaustive events Set of events which includes all possible events is said to be exhaustive. Addition law of probability P(A) and P(B) are probabilities of two events A and B, then P(A È B) = P(A) + P(B) – P(A Ç B) For mutually exclusive events P(A Ç B) = 0. Law of total probability If B1, B2,...... Bn are mutually exclusive events and an event A occurs only if Bi occurs, then :

n

P(A) =

å Pç B

Baye’s Theorem If B1, B2, B3,...... Bn are mutually exclusive and exhaustive random events whose individual probabilities are greater than zero, and A be an event within these events. (P(A) > 0), then,

æB ö Pç i ÷ = èAø

æA Pç è Bi

ö ÷ . P(E i ) ø n æAö å P ç B ÷ P(Bi ) i =1 è i ø

Sample Problem 1. There are two bags A and B having 2 white and 1 black ball in bag A and 1 white and 3 black balls in bag B. A bag is chosen at random and two balls are drawn. The balls were one white and one black ball. What is the probability that it is from bag B ? 1 Probability of choosing a bag = P(A) = P(B) = 2 Let X be probability of choosing 2 balls. We have to find P(B/X) By Baye’s theorem æXö P ç ÷ . P(B) è Bø P(B/X) = æXö æXö P ç ÷ . P(A) + P ç ÷ . P(B) èAø è Bø æXö Pç ÷ 1ö æ èBø ç P(A) = P(B) = ÷ = X X 2ø æ ö æ ö è Pç ÷ + Pç ÷ èAø èBø 2! ´1 2 2 1 C1 ´ C1 1!1! 2 æXö = =1 = Pç ÷ = 3 3 ! 3 3 èAø C2 2 !1! 1

2

æXö Pç ÷ = è Bø

= æ Bö Pç ÷ = èXø

=

3! 3 C1 ´ C1 2! 1 = = 4 4! 3´ 4 C2 2!2! 2 3 1 = 6 2 1 1 2 = 2 2 1 4+3 + 3 2 6 1 6 3 ´ = 2 7 7 1

3

Conditional probability æ Bö P(AB) = P(A) . P ç ÷ èAø æBö P ç ÷ . P(A) æAö èAø Pç ÷ = P(B) è Bø Consider one event B which occurs after occurrence of event A, then P(B Ç A) æ Bö Pç ÷ = P(A) èAø

=

number of elements in A Ç B no of elements in A

æ Bö i.e., P(B Ç A) = P(A) . P ç ÷ èAø [Multiplication theorem of conditional probability]

Sample Problem 1. If a pair of die is thrown and the sum of numbers that appeared is 7. Find the probability that one of dice shows 3. Total elements in sample space = 6 × 6 = 36 Sample spaces of occurred sum of 7, B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} Sample space of occurrence of one element as 3 = [(1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3), (3, 1) (3, 2) (3, 4) (3, 5), (3, 6)] (A Ç B) = {(3, 4), (4, 3)} 6 36 2 P(A Ç B) = 36

P(B) =

P(A) =

11 36

2 P(A Ç B) æAö 36 Pç ÷ = = 6 è Bø P(B) 36 2 1 = = 6 3

Mean, Median, Mode Mean : It is the average or arithmetic mean of observed values.

For normal distribution, x1 + x2 + ...... xn n For frequency distribution, mean f1 x1 + f 2 x2 + ...... + f n xn x = f1 + f 2 + ...... + f n

mean

x =

n

å

fi xi

i =1 n

=

å

fi

i =1

Median : Median is the middle term in the collected data. In case of even number of terms, take average of middle terms. æN ö ç - C÷ è ø´h 2 median = L' + f L-lower limit of median class N-total frequency = S fi h -width of median class. C-cumulative frequency upto class preceding median class f -cumulative frequency of median class Mode : It is the value that occur most frequently or it is the value with maximum frequency.

fi - f i -1 ×h ( fi - f i -1 ) + ( fi - fi +1 ) L-lower limit of class certaining mode fi-maximum frequency fi–1-frequency preceding fi fi+1-frequency just after fi h-size of model class Points to be noted · For asymmetric distribution : mean = medians = mode · For normal distribution : mean – mode = 3(mean – median).

mode =

Sample Problem 1. Find the median and mode of following data. Age group

0-10 10-20 20-30 30-40 40-50

No. of people

5

7

9

Age No. of Cumulative group people frequency 0-10 5 5 10-20 20-30 30-40

7 9 6

12 21 27

40-50

8

35

N = 35 Median class : 20 – 30

6

8

3 æN ö ç -C÷ è ø´h 2 Median = L + f

Probability of r successes, P(r) = where, m = np (constant) m = mean, variance

æ 35 ö ç - 12 ÷ è ø ´10 2 = 20 + 21 = 20.62

Mode Max frequency = a \ Modal class is 20 – 30 \ a = 20 Mode = a +

mr .e - m r!

C( fi - fi -1 ) ( f i - f i -1 ) + ( fi - fi +1 )

= 20 +

10(9 - 7) (9 - 7) + (9 - 6)

= 20 +

10 ´ 2 = 24 2+3

standard deviation = m Normal Distribution A continuous random variable X follow normal distribution if its probability density function f (x) is given by:

f (x) =

1 2ps2

e

-

(X -m )2 2s 2

µ-mean, s-standard deviation. Normal Distribution Curve y or f(x)

Standard Deviation It is the square root of mean of squared of the difference of values from their arithmetic mean. It is denoted by s. S( xi - x )2 Sfi

s=

2

2

S f i di æ Sf d ö -ç i i ÷ ´ C Sf i è Sf i ø

s=

deviation (s2)

Variance : square of standard s coeff. of variation : ´ 100 i.e. ratio of S.D. to mean. x Random Variable In a system, those variables which are one explicit function of time is called random variable. Mean (or Expectation) of random variable X = E(X) E(X) = X(a1) . P(a1) + X(a2) . P(a2) + ...... + X(an) P(an) n

=

å X(ai ) . P(ai ) i =1

Variance of X = s2 = E(X2) – [E(X)]2 Standard deviations = s. Poission Distribution Poission’s distribution is related to probabilities of events which are extremely rare but have large number of individual chances to occur. E.g.- number of persons die by dog-bite in a city. The number of trials n is taken very large and chance of P is taken every small taking np constant.

u

x

For this curve mean, median and mode coincide It is symmetrical about y-axis with equal on both sides. Binomial Distribution Binomial distribution deals with trials of repetitive nature in which occurrence of an event or not is of interest. In a series of trials, the probability of r success in n traits is given by nC pr qn–r r where, p-probability of success q-probability of failure Tips : This is important for GATE exam. Example Probability of zero success = nC0 p0qn–0 = qn Probability of 1 success = nC1 p1qn–1 = npqn–1 Probability of 2 success = nC2 p2qn–2 = n(n – 1) . p2qn–2 etc. Points to be noted Sum of probabilities = qn + nC1 pqn–1 + nC2 p2qn–2 + ...... + pn = (q + p)n = 1 Mean of binomial distribution = np and standard deviation =

npq

4

1.

2.

3.

4.

There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e. each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? (a)

1 2

(b)

1 3

(c)

1 4

(d)

1 5

[2008, 2 marks]

If probability density function of a random variable X is f(x) = x2 for –1 £ x £ 1, and = 0 for any other value of x, then the percentage probability P æç – 1 £ x £ 1 ö÷ is è 3 3ø (a) 0.247 (b) 2.47 [2008, 2 marks] (c) 24.7 (d) 247 A person on a trip has a choice between private car and public transport. The probability of using a private car is 0.45. While using the public transport, further choices available are bus and metro, out of which the probability of commuting by a bus is 0.55. In such a situation, the probability (rounded up to two decimals) of using a car, bus and metro, respectively would be [2008, 2 marks] (a) 0.45, 0.30 and 0.25 (b) 0.45, 0.25 and 0.30 (c) 0.45, 0.55 and 0.00 (d) 0.45, 0.35 and 0.20 The standard normal probability function can be approximated as F(xN) =

(

5.

1

1 + exp –1.7255 x N x N

0.12

)

where, xN = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102 cm is [2009, 2 marks]

6.

7.

8.

(a) 66.7% (b) 50.0% (c) 33.3% (d) 16.7% Two coins are simultaneously tossed. The probability of two heads simultaneously appearing is [2010, 1 mark] (a)

1 8

(b)

1 6

(c)

1 4

(d)

1 2

There are two containers, with one containing 4 red and 3 green balls and the other containing 3 blue and 4 green balls. One ball is drawn at random from each container. The probability that one of the balls is red and the other is blue will be [2011, 1 mark] (a)

1 7

(b)

9 49

(c)

12 49

(d)

3 7

The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000 mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is [2012, 1 mark] (a) < 50% (b) 50% (c) 75% (d) 100% In an experiment, positive and negative values are equally likely to occur. The probability of obtaining at most one negative value in five trials is [2012, 2 marks] (a)

1 32

(b)

2 32

(c)

3 32

(d)

6 32

5

1.

2.

3.

In a city, 40% are women. In an organisation of 5 people from the city, what is the probability that there is exactly two women? (a) 0.4 (b) 0.346 (c) 0.6 (d) 0.426 In a binomial distribution, mean is 5 and variance is 3. Then its mode is (a) 2 (b) 3 (c) 4 (d) 5 What is the probability of having 53 sundays in a leap year? (a)

1 7

(b)

2 7

(a)

10.

(c) 4.

5.

6.

A and B are two elements such that P(A Ç B) = P(A) = (a)

11 40

(b)

8.

21 1 (d) 40 40 A hydraulic structure has four gates which operate independently. The probability of failure of each gate is 0.2. Given that gate 1 has failed, the probability that gate 2 and 3 will fail is (a) 0.24 (b) 0.20 (c) 0.04 (d) 0.008 Six dices are thrown simultaneously. What is the probability that all will show different faces?

(a)

6

4

7!

(b)

4

(d)

14.

15.

5! 64

1 6 6 What is the expectation of number of failures preceeding first success in independent trials with constant probability of success?

(c) 9.

6!

13.

31 40

(c) 7.

12.

1 , 4

2 3 and P(B) = . Then P(A È B) = 5 8

(b)

p +1 q +1

q q +1 (d) p p +1 Let X be a normal random variable with mean 1 and variance 4. The probability P{X < 0} is (a) 0.5 (b) greater than zero and less than 0.5 (c) greater than 0.5 and less than 1.0 (d) 1.0 The probability that a student knows a correct answer in a multiple choice question is 2/3. If the student does not know the answer, he guesses the answer. The probability that the guessed answer being correct is 1/4. Given that the student has answered the question correctly, the conditional probability that the student knows the correct answer is (a) 2/3 (b) 3/4 (c) 5/6 (d) 8/9 An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Ysupplies 40% of the shock absorbers. All shock absorbers are subjected quality test. The ones that pass the quality test are considered reliable. Of X's shock absorber, 96% are reliable. Of Y's shock absorbers, 72% are reliable. The probability that a randomly chosen shock absorber, which is found reliable, is made by Y is A box contains 4 red balls and 6 black balls. Three balls are chosen at random from one box one after other, without replacement. The probability that the selected set contain one red ball and two black balls is (a) 1/20 (b) 1/12 (c) 3/10 (d) 1/2 An unbiased coin is tossed five times. The outcome of each toss either head/tail. Probability of getting at least one head is (a) 1/32 (b) 13/32 (c) 16/32 (d) 31/32 A box contain 2 washers, 3 nuts and 4 bolts. Items are drawn from a box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the forth bolt is: (a) 2/315 (b) 1/630 (c) 1/1260 (d) 1/2520 A standard deviation of a uniformly distributed random variable between 0 and 1 is

(c)

11.

3 4 (d) 7 7 A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. The probability that none of the two screws will be defective (a) 100% (b) 50% (c) 47% (d) none of these For a data, mode is 16 and mean is 22, then median is equal to (a) 60 (b) 66 (c) 20 (d) 38

p q

16.

(a) (c)

1

12

5 12

(b) (d)

1

3

7 12

6 17.

18.

19.

20.

21.

If three coins are tossed simultaneously, the probability of getting at least one head is (a) 1/8 (b) 3/8 (c) 1/2 (d) 7/8 A coin is tossed 4 times. What is the probability of getting heads exactly 3 times? (a) 1/4 (b) 3/8 (c) 1/2 (d) 3/4 Let X and Y be two independent random variables . Which one of the relations between expectations (E), variance (Var) and covariance (Cov) given below is FALSE? (a) E (X Y) = E (X) E (Y) (b) Cov (X, Y) = 0 (c) Var (X + Y) = Var (X) + Var (Y) (d) E (X2Y2) = (E (X))2 + (E(Y))2 A single die is thrown two times, what is the probability that the sum is neither 8 or 9? (a) 1/9 (b) 5/36 (c) 1/4 (d) 3/4 Let U and V be two independent zero mean Gaussian random variables of variances

22.

23.

24.

1 1 and respectively. 9 4

The probability P (3V ³ 2 U) is (a) 4/9 (b) 1/2 (c) 2/3 (d) 5/9 Two independent random variables X and Y are uniformly distributed in the interval [1, 1]. The probability that max [X, Y] is less than. 1/2 is (a) 3/4 (b) 9/16 (c) 1/4 (d) 2/3 A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is (a) 1/3 (b) 1/2 (c) 2/3 (d) 3/4 A fair die is tossed two time. The probability that the 2nd toss results a value that is higher than the first toss is (a)

2 36

(b)

27.

k p (x = k )

28.

29.

30. 31.

32.

26.

(a)

( 2)

(c)

æ1ö ç ÷ è2ø

1

2

(b)

10

æ1ö C2 ç ÷ è2ø

(d)

10

æ1ö C2 ç ÷ è2ø

10

2

10

2 0.2

3 0.4

4 0.2

5 0.1

8 3

(b)

9 3

(c)

17 3

(d)

26

2e 2e 2e 2e3 If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcome are heads?

1 2 1 1 (b) (c) (d) 3 3 4 2 Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability q. what is the probability of a computer being declared faulty? (a) pq + (1 – p) (1 – q) (b) (1 – q) p (c) (1 – p) q (d) pq An exam paper has 150 multiple choice question of 1 mark each with each question having 4 choices. Each incorrect answer fetches – 0.25 marks. Suppose 1000 students chose all their answer randomly with uniform probability. The sum total of the expected marks obtained by all the students is Seven car accidents occurred in a week, what is the probability that they all occurred on the same day?

(a)

33.

(c) 25.

1 0 .1

(a) both student and teacher are right (b) both student and teacher are wrong (c) student is wrong, teacher is right (d) student is right, teacher is wrong Consider two independent random variables X an Y with identical distribution. The variables X and Y take value 0, 1 and 2 with probabilities 1/2, 1/4 and 1/4 resp. What is the conditional probability. P (X + Y = 2/X–Y = 0)? (a) 0 (b) 1/16 (c) 1/6 (d) 1 If E denotes expectation, the variance of the random variable X is given by. (a) E (X2) – E2 (X) (b) E (X2) + E2 (X) 2 (c) E (X ) (d) E2 (X) A fair die is rolled twice. The probability that an odd number will follow an even number is Suppose p is the number of cars per minute passing through a certain road junction between 5 pm and 6 pm, and p has a poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval? (a)

2 6

5 1 (d) 12 2 A coin is tossed independently 4 times. The probability of the event ‘the no. of times heads show up is more than no. of times tail shows up’ is (a) 1/16 (b) 1/8 (c) 1/4 (d) 5/16 A fair coin is tossed 10 times. What is the probability that only the first two tosses will yield heads?

The discrete random variable X takes value from 1 to 5 with probabilities as shown in the table. A student calculates the mean X as 3.5 and her teacher calculates variance to X as 1.5. which of the following statement is true?

34.

35.

(a)

1 7

7

(b)

1 7

6

(c)

1 7

2

(d)

7 27

7

HINTS & SOLUTIONS PAST GATE QUESTIONS EXERCISE 2

1.

(b) P =

6.

(c) Probability of chosing red ball from first container, 4 C1 PR = 7 C1

C1 ´ 25C4 25 C2

Probability of chosing blue ball from second container,

2C ® one selected from 2 defective 1 25C ® four from 25 non-defective 4

Þ P=

2.

(b)

3 C1 PB = 7 C1

1 . 3

Since both these events are mutually independent,

1ö æ 1 Pç – £ x £ ÷ = è 3 3ø

1 3

1 3

ò f ( x ) dx –

1 3

ò

=

x 2 dx

1 3

1 3

= 2ò f ( x ) dx

4 C1 3 C1 12 . = required probability, P = 7 ´ 7 C1 C1 49

7.

(a)

8.

(d) P (positive value) = P (negative value) =

[f(x) is even]

0

1

2 x3 3 = 2× = 81 3

=

0

2 × 100 = 2.47%. 81 (a) P (using private car) = 0.45 \ P (using public transport) = 1 – 0.45 = 0.55 P (using bus) = 0.55 × 0.55 = 0.3 P (using metro) = 1 – (0.45 + 0.3) = 0.25 (d) Probability, P(90 £ x £ 102)

4.

4

4

1.

(b) P(women), p =

40 2 = 100 5

3 5 \ p (2 women members) = 5C2 p2 q3

\ q=

2

1 é 12 ù æ 12 ö 1 + exp ê –1.7255 ç – ÷ – è ø 27 27 úû ë

= 0.165 = 16.5% » 16.7%. (c) Sample space = {(T, T), (H, H), (H, T), (T, H)} (n = 4) Favorable cases = {(H, H)} (m = 1) Probability, P =

1

=

1 1 – = 1 + 1 1 + exp ( 0.6870)

5.

5

PRACTICE EXERCISE

æ 12 ö F(0) – F ç – ÷ è 27 ø 1 – 1 + exp (0)

0

æ 1 ö æ 1ö æ 1ö æ 1ö C0 × ç ÷ ç ÷ + 5C1 ç ÷ ç ÷ è 2ø è 2ø è 2ø è 2ø 5

102 – 102 ö æ 90 – 102 æ –12 ö = Pç £z£ = Pç £ z £ 0÷ ÷ è 27 è 27 ø 27 ø

=

5

1 æ 1ö 6 æ 1ö = 1×1× ç ÷ + 5 × × ç ÷ = . è 2ø 2 è 2ø 32

% of P =

3.

1 2 (Total probability is one and each are equally likely to occur) Required probability = P (no negative value) + P (one negative value)

1 . 4

5 ´ 4 æ 2 ö æ 3ö ç ÷ ç ÷ 1 ´ 2 è 5 ø è 5ø

= .346. 2.

(d) np = 5, npq = 3 \ q=

3 5

\ p=1–

3 2 = 5 5

25 2 Mode x is given by np + p > x > np – q

\ n=

3

8 \ E(x) = 0 . p + 1 . qp + 2 q2p + ..... = qp (1 + 2p + 3q2 + .....)

2 3 5+ >x>5– 5 5

=

27 22 >x> 5 5 5.4 > x > 4.4 \ x = 5.

3.

=

(b) In a leap year, there are 52 full sundays. (Divide 366 by 7) Thus, there are 2 days left P(having sunday) is if days are either saturdaysunday or sunday-monday. Total cases : (S-M), M-T, T-W, W-Th, Th-F, F-Sa, (Sa-S) \ Required probability = 3

4.

(c) P =

10.

11.

(c) 3 × median = mode + 2 × mean = 16 + 2 × 22 = 60 median = 20.

6.

(c) P (A È B) = P(A) + P(B) – P(A Ç B) 12.

=

31 1 – 40 4

=

31 – 10 21 = . 40 40

(c) P = P(2) × P(3) (Since all gates operate independently) = 0.2 × 0.2 = 0.04. (b) Total number of ways of occurance = 65 Number of ways different number occur of the dice = 6!

9.

5

6

=

(b) m = 1 –m ) s

2 3

1 4

æRö æRö P ç ÷ = 0.96 , P ç ÷ = 0.72 èXø è Yø æ Y ö P(Y Ç R) Pç ÷ = èRø P(R)

16 + 15 1 – = 40 4

6!

q . P

=

2 8 3 Required probability = º 2 1 1 9 + ´ 3 3 4

2 3 1 = + – 5 8 4

\ P=

P

2

P (guessed answer is correct) =

5.

8.

qp

1 = P(Z < – ) 2 = P (Z < – 0.5) P (X < 0) = 0.5 – P (0 < z < 5)

7 » 47%. = 15

7.

(1 – q ) 2

P (X < 0) = P (Z <

2 . 7

C0 ´ 7 C2 10 C2

qp

5! 6

4

13.

P(Y)P(R ) Y P(X)P R + P(Y)P R X Y

=

0.4 ´ 0.72 0.6 ´ 0.96 + 0.4 ´ 0.72

=

0.288 = 0.334 .864

( )

(d) P[1R Ç 2B] = =

.

(c) Expectation, E(x) = mean =

=

1 n

14.

n

å xi

i =1

n

=

å Pi xi

i=1

2

3

Probability of success in trials is p, qp, q p, q p, .....

4

( )

C1 ´ 6C1 10

C3

=

4 ´ 15 120

1 2

1 1 (d) n = 5, p = , q = 2 2 P (X ³ 1) = 1 – P (X = 0) æ1ö = 1 – 5C0 ç ÷ è2ø

5

=

31 32

9 15.

(c) Required probability =

16.

2 1 3 2 1 4 3 2 1 1 ´ ´ ´ ´ ´ ´ ´ ´ = 9 8 7 6 5 5 3 2 1 1260

(a) Var (X) =

=

18. 19. 20.

9

Req. prob = 25.

36 =

1

4

= 4 C3 ( 1 2 )

26.

1 1 æ1ö (c) P (first two tosses is head) = ´ = ç ÷ 2 2 è 2ø

21. 22.

= P[X £ x] × P [Y £ y] 1

=

2

ò

–1

1 dx ´ 2

1

2

æ1ö =ç ÷ è2ø

=

8

æ1ö æ1ö 10 Required probability = ç ÷ ´ ç ÷ = ( 1 2 ) è2ø è2ø

27.

(b) m = S x P(x) = (1 × 0.1) + (2 × 0.2) + (3 × 0.4) + (4 × 0.2) + (5 × 0.1) =3 s2 = Sx2P(x) – [SxP(x)]2 = 10.2 – 9 = 1.2 \ both are worng.

28.

(c)

æ X + 4 = 2 ö = P[(X + Y = 2) Ç (X – Y = 0)] Pç ÷ P(X – Y = 0) è X– Y = 0 ø

1

=

ò 2 dy

P(X = 1,Y = 1) (solving) P(X – Y = 0)

–1

=

1 3 3 ´ ´ 4 2 2

= 29. 30.

9 16 3

23.

8

2

\

1 2 1 1/ 2 = [ x ]1/ –1 ´ [ y ]–1 2 2

=

2

1 1 1 P (remaining 8 tosses is tail) = ´ ´ ... ´ (8 times) 2 2 2

\ Required probability (sum not 8 or 9) 1 3 = 4 4 (b) P(3V ³ 2U) = P(3V – 2U ³ 0) Þ P (Z ³ 0) = 1/2 (b) P [max (X, Y)] = P [X £ x, Y £ y] Q X and Y one independent

( 2 ) + 4 C2 ( 1 2 ) 4

3 1

= 5/16

Total 36 cases.

= 1– P = 1–

15 5 = 36 12

(d) n = 4 Let X be event of showing no. of heads. Req. prob = P (X = 3) + P (X = 4)

1

12 (d) Required probability (getting at least one head) = 1 – P (getting no head) = 1 – 3C0 (1/2)0 (1\2)3 = 7/8 (a) P(X = 3) = 4C3 (1/2)3 (1/2) = 1/4 (d) (d) First find probability of getting sum 8 or 9 S = (2, 6) (3, 5) (4, 4) (5, 3) (6, 2); (3, 6) (4, 5) (5, 4) (6, 3) \ n=9

P(E) =

(c) Sample space ; {(1, 2) (1, 3) .... (1, 6); (2, 3) (2, 4) (2, 5) (2, 6); (3, 4) (3, 5), (3, 6); (4, 5) (4, 6); (5, 6)} n = 15

(b – a)2 for a < x < b 12

(1 – 0)2 1 = 12 12

SD = s =

17.

24.

1 4´ 4

æ1 1ö æ1 1ö æ1 1ö ç ´ ÷ +ç ´ ÷+ç ´ ÷ è2 2ø è4 4ø è4 4ø 1

6

(a) Var X = E (X2) – (E(X))2 Ans. 0.25 P (getting odd no.) =

5

1 æ1ö æ1ö (c) Probability = + ç ÷ + ç ÷ + ....¥ 2 è2ø è2ø

1 2 = 2 = 1 3 1– 4

1

P (getting even no.) = \

Req. prob =

1 2

1 2

1 1 1 ´ = = 0.25 2 2 4

10 31.

(c) P(X< 3) = P (X = 0) + P (X = 1) + P (X = 2) = e–3 + 3e–3 +

32.

9e 2

1 0.25

–0.25 0.75

Expected marks = E (X) = S X. P(X) = 1×.25 + (– 0.25) (0.75) = 0.0625 \ Marks for 150 questions of 1000 students = 150 × 1000 × (0.06 25) = 9375

1 3

(a) There can be 2 cases: (1) A faulty computer is declared faulty probability = p × q (2) right computer is declared faulty probability = (1 – p) × (1 – q) \ Required probability = pq + (1 – p) (1 – q)

Ans : 9375 Marks Probability

9 ö 17 –3 –3 æ = e ç1 + 3 + ÷ = e è 2ø 2 (a) Total available out comes ={(H, T) (T, H) (H, H)}

Required probability = 33.

34.

–3

35.

1 1 7 (b) Required probability = C1 ´ 7 = 6 7 7