Probability and Statistics

1

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INTRODUCTIONProbability is the measure of degree of certainity or uncertainityof occurrence of an event. The set of all possible outcomes iscalled 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) £ 1Probability of sample space, P(S) = 1.

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 eventsEvent whose occurrences is independent of occurrence of anotherevents is said to be mutually exclusive. They don’t occursimultaneously i.e. A Ç B = fE.g. : tossing a coin, throwing a die.

Equally likely eventsIf an event cannot happen in performance to other such events issaid to be equally likely.

Independent eventsTwo events are said to be independent if happening or failure ofone does not effect the happening or failure of the others.

Exhaustive eventsSet of events which includes all possible events is said to beexhaustive.

Addition law of probabilityP(A) and P(B) are probabilities of two events A and B, thenP(A È B) = P(A) + P(B) – P(A Ç B)For mutually exclusive events P(A Ç B) = 0.

Law of total probabilityIf B1, B2,...... Bn are mutually exclusive events and an event Aoccurs only if Bi occurs, then :

Probability of occurrence of A is given by

P(A) =1

AP P(B )B

n

iii=

æ öç ÷è ø

åP(A/Bi)-Probability of occurrence of A when Bi has alreadyhappened.

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

BP

Aiæ ö

ç ÷è ø

=

1

AP . P(E )B

AP P(B )B=

æ öç ÷è ø

æ öç ÷è ø

å

ii

n

iii

Sample Problem1. 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 ischosen at random and two balls are drawn. The balls wereone white and one black ball. What is the probability that itis from bag B ?

Probability of choosing a bag = P(A) = P(B) = 12

Let X be probability of choosing 2 balls.We have to find P(B/X)By Baye’s theorem

P(B/X) =

XP . P(B)B

X XP . P(A) + P . P(B)A B

æ öç ÷è ø

æ ö æ öç ÷ ç ÷è ø è ø

=

XP 1B P(A) = P(B) =X X 2P PA B

æ öç ÷ æ öè ø

ç ÷è øæ ö æ ö+ç ÷ ç ÷

è ø è ø

XPA

æ öç ÷è ø

=2 1

1 13

2

2 ! 21C C 1!1! 1

3! 3C2 !1! 1

´´

= = =23

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PROBABILITY AND STATISTICS

2


XPB

æ öç ÷è ø

=1 3

1 14

2

3! 31C C 2 ! 1

4 ! 3 4C2 ! 2 ! 2

´´

= =´

=36

=12

BPX

æ öç ÷è ø

=

1 12 2

2 1 4 33 2 6

=+

+

=1 62 7

´ =37

Conditional probability

P(AB) = P(A) . BPA

æ öç ÷è ø

APB

æ öç ÷è ø

=

BP . P(A)AP(B)

æ öç ÷è ø

Consider one event B which occurs after occurrence of event A,then

BPA

æ öç ÷è ø

=P(B A)

P(A)Ç

=number of elements in A B

no of elements in AÇ

i.e., P(B Ç A) = P(A) . BPA

æ öç ÷è ø

[Multiplication theorem of conditional probability]

Sample Problem1. 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 = 36Sample 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)}

P(B) =636

P(A) = 1136

P(A Ç B) =2

36

APB

æ öç ÷è ø

=P(A B)

P(B)Ç

=

2366

36

=26

=13

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

For normal distribution,

mean x = 1 2 ...... nx x xn

+ +

For frequency distribution, mean

x = 1 1 2 2

1 2

............

n n

n

f x f x f xf f f+ + +

+ + +

= 1

1

n

in

i

fi xi

fi

=

=

å

åMedian : Median is the middle term in the collected data. In caseof even number of terms, take average of middle terms.

median =

N C2L' +

æ ö-ç ÷è ø ´ h

fL-lower limit of median classN-total frequency = S fih -width of median class.C-cumulative frequency upto class preceding median classf -cumulative frequency of median class

Mode : It is the value that occur most frequently or it is the valuewith maximum frequency.

mode = 1

1 1( ) ( )i i

i i i i

f ff f f f

-

- +

-- + -

× h

L-lower limit of class certaining modefi-maximum frequencyfi–1-frequency preceding fifi+1-frequency just after fih-size of model classPoints to be noted· For asymmetric distribution : mean = medians = mode· For normal distribution : mean – mode

= 3(mean – median).

Sample Problem1. Find the median and mode of following data.

Age group 0-10 10-20 20-30 30-40 40-50No. of people 5 7 9 6 8

Age No. of Cumulativegroup people frequency0-10 5 5

10-20 7 1220-30 9 2130-40 6 2740-50 8 35

N = 35Median class : 20 – 30

3


Median =

N C2L h

f

æ ö-ç ÷è ø+ ´

=

35 12220 10

21

æ ö-ç ÷è ø+ ´

= 20.62ModeMax frequency = a\ Modal class is 20 – 30 \ a = 20

Mode = 1

1 1

C( )( ) ( )

i i

i i i i

f fa

f f f f-

- +

-+

- + -

= 10(9 7)20(9 7) (9 6)

-+

- + -

=10 2202 3

´+ =

+24

Standard DeviationIt is the square root of mean of squared of the difference of valuesfrom their arithmetic mean. It is denoted by s.

s =2( )i

i

x xf

S -S

s =22

CS Sæ ö- ´ç ÷S Sè ø

i i i i

i i

f d f df f

Variance : square of standard deviation (s2)

coeff. of variation : 100xs

´ i.e. ratio of S.D. to mean.

Random VariableIn a system, those variables which are one explicit function oftime 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)

=1

X( ) . P( )n

i ii

a a=å

Variance of X = s2 = E(X2) – [E(X)]2

Standard deviations = s.Poission DistributionPoission’s distribution is related to probabilities of events whichare extremely rare but have large number of individual chances tooccur. E.g.- number of persons die by dog-bite in a city. Thenumber of trials n is taken very large and chance of P is takenevery small taking np constant.

Probability of r successes, P(r) = .!

r mm er

-

where, m = np (constant)m = mean, variance

standard deviation = mNormal DistributionA continuous random variable X follow normal distribution if itsprobability density function f (x) is given by:

f (x) =

2

2(X )

22

1

2

-m-

s

pse

µ-mean, s-standard deviation.Normal Distribution Curve

y or f(x)

u x

For this curve mean, median and mode coincideIt is symmetrical about y-axis with equal on both sides.

Binomial DistributionBinomial distribution deals with trials of repetitive nature in whichoccurrence of an event or not is of interest.In a series of trials, the probability of r success in n traits is givenby

nCr pr qn–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 notedSum 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. There are 25 calculators in a box. Two of them aredefective. Suppose 5 calculators are randomly picked forinspection (i.e. each has the same chance of beingselected), what is the probability that only one of thedefective calculators will be included in the inspection?

(a)12

(b)13

[2008, 2 marks]

(c)14

(d)15

2. If probability density function of a random variable X isf(x) = x2 for –1 £ x £ 1, and

= 0 for any other value of x,

then the percentage probability 1 1P – x

3 3æ ö£ £ç ÷è ø

is

(a) 0.247 (b) 2.47 [2008, 2 marks](c) 24.7 (d) 247

3. A person on a trip has a choice between private car andpublic transport. The probability of using a private caris 0.45. While using the public transport, further choicesavailable are bus and metro, out of which the probabilityof commuting by a bus is 0.55. In such a situation, theprobability (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

4. The standard normal probability function can beapproximated as

F(xN) = ( )0.12N N

1

1 exp –1.7255 x x+

where, xN = standard normal deviate. If mean andstandard deviation of annual precipitation are 102 cm and27 cm respectively, the probability that the annualprecipitation will be between 90 cm and 102 cm is

[2009, 2 marks]

(a) 66.7% (b) 50.0%(c) 33.3% (d) 16.7%

5. Two coins are simultaneously tossed. The probability oftwo heads simultaneously appearing is [2010, 1 mark]

(a)18

(b)16

(c)14

(d)12

6. There are two containers, with one containing 4 red and3 green balls and the other containing 3 blue and 4 greenballs. One ball is drawn at random from each container.The probability that one of the balls is red and the otheris blue will be [2011, 1 mark]

(a)17

(b)949

(c)1249

(d)37

7. The annual precipitation data of a city is normallydistributed with mean and standard deviation as 1000 mmand 200 mm, respectively. The probability that the annualprecipitation will be more than 1200 mm is [2012, 1 mark](a) < 50% (b) 50%(c) 75% (d) 100%

8. In an experiment, positive and negative values are equallylikely to occur. The probability of obtaining at most onenegative value in five trials is [2012, 2 marks]

(a)132

(b)2

32

(c)3

32(d)

632

5


1. In a city, 40% are women. In an organisation of 5 peoplefrom the city, what is the probability that there is exactlytwo women?(a) 0.4 (b) 0.346(c) 0.6 (d) 0.426

2. In a binomial distribution, mean is 5 and variance is 3.Then its mode is(a) 2 (b) 3(c) 4 (d) 5

3. What is the probability of having 53 sundays in a leapyear?

(a)17

(b)27

(c)37

(d)47

4. 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 willbe defective(a) 100% (b) 50%(c) 47% (d) none of these

5. For a data, mode is 16 and mean is 22, then median isequal to(a) 60 (b) 66(c) 20 (d) 38

6. A and B are two elements such that P(A Ç B) = 1 ,4

P(A) =25

and P(B) = 3.8

Then P(A È B) =

(a)1140

(b)3140

(c)2140

(d)140

7. A hydraulic structure has four gates which operateindependently. The probability of failure of each gate is0.2. Given that gate 1 has failed, the probability that gate2 and 3 will fail is(a) 0.24 (b) 0.20(c) 0.04 (d) 0.008

8. Six dices are thrown simultaneously. What is theprobability that all will show different faces?

(a) 46!6

(b) 45!6

(c) 47!6

(d)16

9. What is the expectation of number of failures preceedingfirst success in independent trials with constantprobability of success?

(a) pq

(b) p 1q 1

++

(c) qp

(d) q 1p 1

++

10. Let X be a normal random variable with mean 1 andvariance 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

11. The probability that a student knows a correct answer ina multiple choice question is 2/3. If the student does notknow the answer, he guesses the answer. The probabilitythat the guessed answer being correct is 1/4. Given thatthe student has answered the question correctly, theconditional probability that the student knows the correctanswer is(a) 2/3 (b) 3/4(c) 5/6 (d) 8/9

12. An automobile plant contracted to buy shock absorbersfrom two suppliers X and Y. X supplies 60% and Ysupplies40% of the shock absorbers. All shock absorbers aresubjected quality test. The ones that pass the quality testare considered reliable. Of X's shock absorber, 96% arereliable. Of Y's shock absorbers, 72% are reliable. Theprobability that a randomly chosen shock absorber, whichis found reliable, is made by Y is

13. A box contains 4 red balls and 6 black balls. Three ballsare chosen at random from one box one after other, withoutreplacement. The probability that the selected set containone red ball and two black balls is(a) 1/20 (b) 1/12(c) 3/10 (d) 1/2

14. An unbiased coin is tossed five times. The outcome ofeach toss either head/tail. Probability of getting at leastone head is(a) 1/32 (b) 13/32(c) 16/32 (d) 31/32

15. A box contain 2 washers, 3 nuts and 4 bolts. Items aredrawn from a box at random one at a time withoutreplacement. The probability of drawing 2 washers firstfollowed by 3 nuts and subsequently the forth bolt is:(a) 2/315 (b) 1/630(c) 1/1260 (d) 1/2520

16. A standard deviation of a uniformly distributed randomvariable between 0 and 1 is

(a) 112 (b) 1

3

(c)512

(d)712

6


17. If three coins are tossed simultaneously, the probabilityof getting at least one head is(a) 1/8 (b) 3/8(c) 1/2 (d) 7/8

18. A coin is tossed 4 times. What is the probability ofgetting heads exactly 3 times?(a) 1/4 (b) 3/8(c) 1/2 (d) 3/4

19. Let X and Y be two independent random variables . Whichone 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

20. A single die is thrown two times, what is the probabilitythat the sum is neither 8 or 9?(a) 1/9 (b) 5/36(c) 1/4 (d) 3/4

21. Let U and V be two independent zero mean Gaussian

random variables of variances 14

and 19 respectively.

The probability P (3V ³ 2 U) is(a) 4/9 (b) 1/2(c) 2/3 (d) 5/9

22. Two independent random variables X and Y are uniformlydistributed in the interval [1, 1]. The probability thatmax [X, Y] is less than. 1/2 is(a) 3/4 (b) 9/16(c) 1/4 (d) 2/3

23. 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

24. A fair die is tossed two time. The probability that the 2ndtoss results a value that is higher than the first toss is

(a)2

36 (b)26

(c)5

12(d)

12

25. A coin is tossed independently 4 times. The probabilityof the event ‘the no. of times heads show up is more thanno. of times tail shows up’ is(a) 1/16 (b) 1/8(c) 1/4 (d) 5/16

26. A fair coin is tossed 10 times. What is the probability thatonly the first two tosses will yield heads?

(a) ( )212 (b)

210

21C2

æ öç ÷è ø

(c)101

2æ öç ÷è ø

(d)10

102

1C2

æ öç ÷è ø

27. The discrete random variable X takes value from 1 to 5with probabilities as shown in the table. A studentcalculates the mean X as 3.5 and her teacher calculatesvariance to X as 1.5. which of the following statement istrue?

k 1 2 3 4 5p ( x = k ) 0 .1 0.2 0.4 0.2 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

28. Consider two independent random variables X an Y withidentical distribution. The variables X and Y take value 0,1 and 2 with probabilities 1/2, 1/4 and 1/4 resp. What isthe conditional probability. P (X + Y = 2/X–Y = 0)?(a) 0 (b) 1/16(c) 1/6 (d) 1

29. If E denotes expectation, the variance of the randomvariable X is given by.(a) E (X2) – E2 (X) (b) E (X2) + E2 (X)(c) E (X2) (d) E2 (X)

30. A fair die is rolled twice. The probability that an oddnumber will follow an even number is

31. Suppose p is the number of cars per minute passingthrough a certain road junction between 5 pm and 6 pm,and p has a poisson distribution with mean 3. What is theprobability of observing fewer than 3 cars during anygiven minute in this interval?

(a) 38

2e(b) 3

92e

(c) 3172e

(d) 3262e

32. If two fair coins are flipped and at least one of theoutcomes is known to be a head, what is the probabilitythat both outcome are heads?

(a)13 (b)

14

(c)12

(d)23

33. Consider a company that assembles computers. Theprobability of a faulty assembly of any computer is p. Thecompany therefore subjects each computer to a testingprocess. This testing process gives the correct result forany computer with a probability q. what is the probabilityof a computer being declared faulty?(a) pq + (1 – p) (1 – q) (b) (1 – q) p(c) (1 – p) q (d) pq

34. An exam paper has 150 multiple choice question of 1 markeach with each question having 4 choices. Each incorrectanswer fetches – 0.25 marks. Suppose 1000 students choseall their answer randomly with uniform probability. Thesum total of the expected marks obtained by all thestudents is

35. Seven car accidents occurred in a week, what is theprobability that they all occurred on the same day?

(a) 717

(b) 61

7(c) 7

12

(d) 772

7


PAST GATE QUESTIONS EXERCISE

1. (b) P = 2 25

1 425

2

C CC

´

2C1 ® one selected from 2 defective25C4 ® four from 25 non-defective

Þ P = 1

.3

2. (b) ( )

13

1–3

1 1P – x f x dx3 3

æ ö£ £ =ç ÷è ø ò =

13

2

1–3

x dxò

= ( )

13

02 f x dxò [f(x) is even]

=

13 3

0

x23

× = 281

% of P = 281

× 100 = 2.47%.

3. (a) P (using private car) = 0.45\ P (using public transport) = 1 – 0.45 = 0.55P (using bus) = 0.55 × 0.55 = 0.3P (using metro) = 1 – (0.45 + 0.3) = 0.25

4. (d) Probability, P(90 £ x £ 102)

= 90 – 102 102 – 102

P z27 27

æ ö£ £ç ÷è ø = –12

P z 027

æ ö£ £ç ÷è ø

F(0) – 12F –

27æ öç ÷è ø

= 1 1–

1 exp (0) 12 121 exp –1.7255 – –27 27

+ é ùæ ö+ ç ÷ê úè øë û

= ( )1 1–

1 1 1 exp 0.6870+ +

= 0.165 = 16.5% » 16.7%.5. (c) Sample space = {(T, T), (H, H), (H, T), (T, H)}

(n = 4)Favorable cases = {(H, H)} (m = 1)

Probability, P = 1

.4

6. (c) Probability of chosing red ball from first container,

PR = 4

17

1

CC

Probability of chosing blue ball from second container,

PB = 3

17

1

CC

Since both these events are mutually independent,

required probability, P = 4 3

1 17 7

1 1

C C 12 .49C C

´ =

7. (a)

8. (d) P (positive value) = P (negative value) = 12

(Total probability is one and each are equally likelyto occur)Required probability

= P (no negative value) + P (one negative value)

= 0 5 1 4

5 50 1

1 1 1 1C C

2 2 2 2æ ö æ ö æ ö æ ö× +ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø

= 5 41 1 1 61 1 5 .

2 2 2 32æ ö æ ö× × + × × =ç ÷ ç ÷è ø è ø

PRACTICE EXERCISE

1. (b) P(women), p = 40 2

100 5=

\ q = 35

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

= 2 35 4 2 3

1 2 5 5´ æ ö æ ö

ç ÷ ç ÷è ø è ø´

= .346.2. (d) np = 5, npq = 3

\ q = 35

\ p = 1 – 3 25 5

=

\ n = 252

Mode x is given bynp + p > x > np – q

HINTS & SOLUTIONS

8


5 + 25

> x > 5 – 35

27 22x

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 leftP(having sunday) is if days are either saturday-sunday or sunday-monday.Total cases : (S-M), M-T, T-W, W-Th, Th-F,F-Sa, (Sa-S)

\ Required probability = 2

.7

4. (c) P = 3 7

0 210

2

C C

C

´

= 7

15 » 47%.

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

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

= 2 3 1–5 8 4

+

= 16 15 1–

40 4+

= 31 1

–40 4

= 31 – 10 21.

40 40=

7. (c) P = P(2) × P(3)(Since all gates operate independently)

= 0.2 × 0.2 = 0.04.

8. (b) Total number of ways of occurance = 65

Number of ways different number occur ofthe dice = 6!

\ P = 5 46! 5!

.6 6

=

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

= n

ii 1

1 xn =

å = n

i ii 1

P x=å

Probability of success in trials is p, qp, q2p, q3p, .....

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

= ( )2

qp

1 – q

= 2qp q .

PP=

10. (b) m = 1

P (X < 0) = P (Z < –ms

)

1P(Z – )2

= <

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

11. (d) P(answering correctly) 23

=

P (guessed answer is correct) 14

=

Required probability

283

2 1 1 93 3 4

= º+ ´

12.RP 0.96X

æ ö =ç ÷è ø

, RP 0.72Y

æ ö =ç ÷è ø

Y P(Y R)PR P(R)

Çæ ö =ç ÷è ø

( ) ( )RP(Y)P( )Y

R RP(X)P P(Y)PX Y=

+

0.4 0.72

0.6 0.96 0.4 0.72´

=´ + ´

0.288.864

= = 0.334

13. (d) P[1R 2B]Ç 4 6

1 110

3

C C 4 15120C

´ ´= =

12

=

14. (d) n = 5, 1 1,2 2

p q= =

P (X ³ 1) = 1 – P (X = 0)

5

50

11– C2

æ ö= ç ÷è ø

3132

=

9


15. (c) Required probability

2 1 3 2 1 4 39 8 7 6 5 5 3

= ´ ´ ´ ´ ´ ´ 2 1 12 1 1260

´ ´ =

16. (a) Var (X) = 2( – )

12b a

for a < x < b

2(1– 0) 1

12 12= =

SD = 112

s =

17. (d) Required probability (getting at least one head)= 1 – P (getting no head)= 1 – 3C0 (1/2)0 (1\2)3 = 7/8

18. (a) P(X = 3) = 4C3 (1/2)3 (1/2) = 1/419. (d)20. (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

9 136 4P(E) = = Total 36 cases.

\ Required probability (sum not 8 or 9)

= 34

11 – P 1–4

= =

21. (b) P(3V ³ 2U) = P(3V – 2U ³ 0)Þ P (Z ³ 0) = 1/2

22. (b) P [max (X, Y)] = P [X £ x, Y £ y]Q X and Y one independent

= P[X £ x] × P [Y £ y]

12

12

–1 –1

1 12 2

dx dy= ´ò ò

1/ 2 1/ 2–1 –1

1 1[ ] [ ]2 2

x y= ´

1 3 34 2 2

= ´ ´

916

=

23. (c) Probability 3 51 1 1 ....

2 2 2æ ö æ ö= + + + ¥ç ÷ ç ÷è ø è ø

122

1 31–4

= =

24. (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

Req. prob = 15 536 12

=

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

( ) ( ) ( )3 44 41 1 13 22 2 2= C + C

= 5/16

26. (c) P (first two tosses is head) 1 1=2 2

´ 21

2æ ö= ç ÷è ø

P (remaining 8 tosses is tail) 1 1 1...2 2 2

= ´ ´ ´ (8 times)

81

2æ ö= ç ÷è ø

\ Required probability 2 81 1

2 2æ ö æ ö= ´ç ÷ ç ÷è ø è ø ( )101

2=

27. (b) m = S x P(x) = (1 × 0.1) + (2 × 0.2) + (3 × 0.4)+ (4 × 0.2) + (5 × 0.1)

= 3s2 = Sx2P(x) – [SxP(x)]2

= 10.2 – 9 = 1.2\ both are worng.

28. (c)X 4 2PX– Y 0

+ =æ öç ÷=è ø

P[(X Y 2) (X – Y 0)]P(X – Y 0)

+ = Ç ==

=

P(X 1,Y 1)P(X – Y 0)

= ==

= (solving)

1 14 4

1 1 1 1 1 12 2 4 4 4 4

´=

æ ö æ ö æ ö´ + ´ + ´ç ÷ ç ÷ ç ÷è ø è ø è ø

16=

29. (a) Var X = E (X2) – (E(X))2

30. Ans. 0.25

P (getting odd no.) 12

=

P (getting even no.) 12

=

\ Req. prob 1 1 1 0.252 2 4

= ´ = =

10


31. (c) P(X< 3) = P (X = 0) + P (X = 1) + P (X = 2)

= e–3 + 3e–3 + –39

2e

= –3 –39 171 32 2

e eæ ö+ + =ç ÷è ø

32. (a) Total available out comes ={(H, T) (T, H) (H, H)}

Required probability 13

=

33. (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)

34. Ans : 9375

Marks 1 –0.25Probability 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

35. (b) Required probability 7

1 7 61 1C

7 7= ´ =

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