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**Q.1. An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then the expected value of X, is (2020)(1) 3/16(2) 1/8(3) -3/16(4) -1/8Ans.** (2)

We have

The expected value of X = ∑ Xp(k)

(1) 17/8

(2) 17/4

(3) 17/2

(4) 4

Ans.

The probability of machine being faulty = 1/4

The probability of machine not being faulty

Hence, the probability that at most two machines will be out of service on the same day is

Thus,

(1) P (A/B) = 2/3

(2) P(A/B') = 1/3

(3) P(A'/B') = 1/3

(4) P(A/A ∪ B) = 1/4

Ans.

Given, A and B are two independent events, then

P(A/B) = P(A) = 1/3

P(A/B') = P(A)= 1/3

P(A'/B') = P(A') = 1 - P(A)

(1) 965/2

(2) 965/2

(3) 945/2

(4) 945/2

Ans.

(1) 7/12

(2) 1/36

(3) 1/6

(4) 23/36

Ans.

We have

⇒ 6K

⇒ K = 1/6 or K = -1(rejected)

Now, P (X > 2) = K + 2K + 5K

(1) 0.02

(2) 0.20

(3) 0.01

(4) 0.10

Ans.

Given,

P(A) + P(B) - 2P(A∩B) = 2/5 ...(1)

and P(A∪B) = 1/2

⇒ P(A) + P(B) - P(A∩B) = 1/2 ...(2)

From Eqs. (1) and (2), we get

Hence, the probability of both events occur together is 0.10.

(1) 9/16

(2) 11/16

(3) 13/16

(4) 15/16

Ans.

The required probability is

(1) 21/49

(2) 27/49

(3) 26/49

(4) 32/49

Ans.

Let G represents drawing a green ball and R represents drawing a red ball

So, the probability that second drawn ball is red

= 32/49

(1) 13/36

(2) 15/72

(3) 19/72

(4) 19/36

Ans.

p(Outcome is head) = 1/2

P(Outcome is tail) = 1/2

P( 7 or 8 is the sum of two dice) = = 11/36

P(7 or 8 is the number of card)

Required probability =

(1) 3

(2) 6

(3) 5

(4) 4

Let the number of independent shots required to hit the target at least once be n, then

Hence, the above inequality holds when least value of n is 5.

(1) 7/10

(2) 1/2

(3) 2/5

(4) 3/5

Ans.

Probability of getting sum of selected two numbers is even

Probability of getting sum is even and selected numbers are also even

Hence,

(1) 7/2

(2) 5/2

(3) 4/2

(4) 6/2

Ans.

Since total number of subsets of the set S = 2

Now, the sum of all number from 1 to 20 =

Then, find the sets which has sum 7.

(1) {7}

(2) {1,6}

(3) {2,5}

(4) {3,4}

(5) {1,2,4}

Then, there is only 5 sets which has sum 203

Hence required probability = 5/2

(1) 200/6

(2) 150/6

(3) 225/6

(4) 175/6

Ans.

Since, the experiment will end in the fifth throw.

Hence, the possibilities are 4**4 4, *4*4 4, ***4 4

(where * is any number except 4)

Required Probability

(1) 400/9 loss

(2) 0

(3) 400/3 gain

(4) 400/3 loss

Ans.

Probability of getting 5 or 6 = P(E) = 2/6 = 1/3

Probability of not getting 5 or 6 = P(E) =

E will consider as success.

His expected gain/loss

(1) 1/6

(2) 1/3

(3) 2/3

(4) 5/6

Ans.

P = Set of students who opted for NCC

Q = Set of Students who opted for NSS

= 40 + 30-20 = 50

∴ Hence, required probability = = 1/6

(1) P(A|B) = P(B) - P(A)

(2) P(A|B) ≥ P(A)

(3) P(A|B) ≤ P(A)

(4) P(A|B) = 1

Ans.

Q.17. The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is: (2019)

(1) 5

(2) 3

(3) 4

(4) 2

Ans.

Let, p is probability for getting head and is probability for getting tail.

(1) 25/192

(2) 7/32

(3) 1/192

(4) 25/32

Ans.

P (at least one hit the target) = 1 - P (none of them hit the target)

(1) 1/11

(2) 1/10

(3) 1/12

(4) 1/17

Ans.

Let, A = At least two girls

B = All girls

(1) 5

(2) 6

(3) 8

(4) 7

Ans.

According to the question,

Hence, minimum value is 7.

(1) 1/10

(2) 1/5

(3) 3/10

(4) 3/20

Ans.

Total no. of triangles =

Favorable no. of triangle i.e, equilateral triangles (ΔAEC and ΔBDF) = 2.

Hence, required probability = 1/10

(1) 17

(2) 121

(3) 1

(4) 137

Ans.

Given mean μ = 8 and variance σ

⇒ μ = np = 8 and σ

(1)

(2)

(3)

(4)

Ans.

Let p is the probability that candidate can solve a problem and q is the probability that candidate can not not solve a problem.

(∵ p + q = 1)

Probability of solving either 50 or 49 problem by the candidate

(1) 1/2 gain

(2) 1/4 loss

(3) 1/2 loss

(4) 2 gain

Ans.

Let X be the random variable which denotes the Rs gained by the person.

Total cases = 6 x 6 = 36.

Favorable cases for the person on winning ₹ 15 are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) i.e., 6 cases.

Favorable cases for the person on winning ₹ 12 are (6, 3), (5, 4), (4, 5), (3, 6) i.e., 4.

Remaining cases = 36 - 6 - 4 = 26

(1) 3/10

(2) 2/5

(3) 1/5

(4) 3/4

Initially 4 Red balls and 6-Black balls

Required probability =

(1) 9/16

(2) 9/32

(3) 7/8

(4) 7/16

Ans.

Probability that box A is selected P(A) = 1/2

Probability that box B is selected P(B) = 1/2

E be event that one ball is white while the other is red

P(E) = P(A) . P(E/A) + P(B) P(E/B)

(1) 6

(2) 5

(3) 3

(4) 4

Ans.

Let the number of children in each family be x. Thus the total number of children in both the families are 2x.

Now, it is given that 3 tickets are distributed amongst the children of these two families.

Thus, the probability that all the three tickets go to the children in family B.

Thus, the number of children in each family is 5.

(1)

(2)

(3)

(4)

(1) 6/25

(2) 12/5

(3) 6

(4) 4

Ans.

n = 10

p(Probability of drawing a green ball) = 15/25

(1) 7/55

(2) 6/55

(3) 12/55

(4) 14/55

Ans.

Total number of ways =

Favourable ways are

(0, 4), (0, 8), (4, 8), (2, 6), (2, 10), (6, 10)

Probability = 6/55

(1) 3/16

(2) 7/32

(3) 7/16

(4) 7/64

Ans.

(1) 39/64

(2) 21/64

(3) 9/64

(4) 15/64

Ans.

= 27/64

(1) 127/128

(2) 63/64

(3) 255/256

(4) 1/2

Ans.

(1) 3/11

(2) 2/23

(3) 1/11

(4) 21/220

Ans.

Probability of 4 member committee which contain atleast one woman.

Probability of committees to have more women than men.

(1) 4/3

(2) 1/3

(3) 3/2

(4) 5/12

Ans.

(1) E

(2) E

(3) E

(4) E

Ans.

P(E

P(E

P(E

P(E

P(E

P(E

P(E

∴ E

(1) 8/17

(2) 1/4

(3) 5/17

(4) 11/20

Ans.

P(A) = 2/5

(1) 192/729

(2) 256/729

(3) 240/729

(4) 496/729

Ans.

Given = p = 2q & we know that p+q = 1

⇒ P = 2/3, q = 1/3

The problem of at least 5 successes

=

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