Practice question on probability for iit jee , NDA , and Airforce page no. 2

Probability

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Question no. 16

The mean and variance of a binomial distribution are 4 and 3 respectively , then he probability of getting six successes in this distribution , is

looks_one

looks_two

looks_3

looks_4

Option looks_two

Solution :
np = 4 ....[1]
npq = 3 ....[2]
divide [2] by [1]
q = 3/4
p + q = 1
p = 1/4
n*(1/4) = 4
n = 16
P(six success) = ¹⁶C₆(1/4)⁶(3/4)¹⁰

Question no. 17

A coin is tossed 10 time. The probability of getting exactly six heads is

looks_one

looks_two

looks_3

looks_4 none of these

Option looks_two

Solution :
n=10
p = q = 1/2
Probability of getting exactly six heads = ¹⁰C₆(1/2)⁶(1/2)⁴ = 105/512

Question no. 18

A five digit number is written down at random. The probability that the number is divisible by 5 and no two consecutive digits are identical, is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_3

Solution :
Total number of ways = 9*10*10*10*10
Number should be divisible by 5 , so last digit must be 0 , 5
Number can be written at first digit = 9
Number can be written at second digit = 9
Number can be written at third digit = 9
Number can be written at forth digit = 8
Number of ways = 9*9*9*9*8
probability that the number is divisible by 5 and no two consecutive digits are identical = (9*9*9*9*8)/9*10*10*10*10 = (3/5)⁵

Question no. 19

Fifteen coupons are numbered 1 to 15. Seven coupons ae selected at random, one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9 , is

looks_one

looks_two

looks_3

looks_4 none of these

Option looks_one

Solution :
Probability that the largest number appearing on a coupon is 9 = 9/15 = 3/5
Probability that the largest number appearing on 7 coupons is 9 = (3/5)⁷

Question no. 20

A fair dice is tossed eight times. The probability that a third six is observed in the eight throw is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_two

Solution :
till the seventh throw , there must be 2 throw of six. So
probability of having 2 throw at 7
n = 7
p = 1/6
q = 5/6
P = ⁷C₂(1/6)²(5/6)²
probability of having six at one throw = 1/6
The probability that a third six is observed in the eight throw = ⁷C₂(1/6)²(5/6)² * (1/6) = ⁷C₂ (5)⁵(1/6)⁸

Question no. 21

If X follows a binomial distribution with parameters n =100 and p= 1/3 then P(X=r) is maximum when r =

looks_one 32

looks_two 34

looks_3 33

looks_4 31

Option looks_3 33

solution :
P(X=r) is maximum
then r = (n+1)p
p= ( 100 + 1 )(1/3) = 33.3 = 33

Question no. 22

If X follows a binomial distribution with parameters n = 8 and p = 1/2, then P(|X-4|≤ 2) equals

looks_one

looks_two

looks_3

looks_4 none of these

Option looks_two

solution :
|X-4|≤ 2
Case 1 : X - 4 ≤ 2
X ≤ 6
Case 2 : -(X - 4 ) ≤ 2
-X + 4 ≤ 2
2 ≤ X
So X : 2 ≤X ≤6
P(2 ≤X ≤6) = P(x=2) + P(x=3) + P(x=4) +P(x=5) +P(x=6)
P(2 ≤X ≤6) = ⁸C₂(1/2)⁸ + ⁸C₃(1/2)⁸ + ⁸C₄(1/2)⁸ + ⁸C₅(1/2)⁸ +⁸C₆(1/2)⁸
P = 119/128

Question no. 23

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

looks_one 2/3

looks_two 4/5

looks_3 7/8

looks_4 15/16

Option looks_4 15/16

Solution :
np = 2 ......[1]
npq = 1 ....[2]
from [1] and [2]
q = 1/2 , p = 1/2 , n = 4
P(X ≥ 1 ) = 1 - ⁴C₁(1/2)⁴ = 15/16

Question no. 24

The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8 is

looks_one 7

looks_two 6

looks_3 5

looks_4 3

Option looks_4 3

Solution :
0.8 = ⁿC₁(1/2)(1/2)^n-1 + ⁿC₂(1/2)²(1/2)^n-2 + .........+ ⁿC_n(1/2)ⁿ
0.8(2ⁿ) = ⁿC₁ + ⁿC₂ + ...... + ⁿC_n
0.8(2ⁿ) = 2ⁿ - ⁿC₀
0.2(2ⁿ) = 1
2ⁿ = 5
n = 2.3 ≃ 3

Question no. 25

A fair coin is tossed 99 times. If X is the number of times heads occur, then P (X=r) is maximum when r is

looks_one 49,50

looks_two 50,51

looks_3 51, 52

looks_4 None of these

Option looks_one 49,50

Solution :
n = 99
For P(x=r) maximum , m = (n+1)p , m-1 are the value of r
p = 1/2
m =(99+1)*(1/2) = 50
m-1 = 50- 1 = 49

Question no. 26

One hundred identical coins, each with probability p of showing heads are tossed once. If 0 > (p) < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins , the value of p is

looks_one 1/2

looks_two 51/101

looks_3 49/101

looks_4 None of these

Option looks_two 51/101

Solution :
¹⁰⁰C₅₀(p)⁵⁰ (1-p)⁵⁰ = ¹⁰⁰C₅₁(p)⁵¹ (1-p)⁴⁹
(1-p)/50 = p/51
p = 51/101

Question no. 27

Let X denote the number of times head occur in n tosses of fair coin. If P(X = 4), P(X= 5) and P(X=6) are in AP ; the value of n is

looks_one 7, 14

looks_two 10, 14

looks_3 12 ,7

looks_4 14, 12

Option looks_one 7, 14

Solution :
2P(X=5) = P(X=4) + P(X=6)
2ⁿC₅ = ⁿC₄ + ⁿC₆

Question no. 28

If X is a binomial variate with parameters n and p where 0 > (p) < 1 such that is independent of n and r ,then p equals

looks_one 1/2

looks_two 1/3

looks_3 1/4

looks_4 None of these

Option looks_one 1/2

solution :
P(X= r) = ⁿC_r(p)^r(q)^n-r
P(X= n-r) = ⁿC_n-r(p)^n-r(q)^r

Question no. 29

A fair die is thrown twenty times. The probability that on the tenth throw the fourth Six appears is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_3

solution :
P(r=3) = ⁹C₃(1/6)³(5/6)⁶
P(having six at thrown) =1 / 6
probability that on the tenth throw the fourth Six appears = ⁹C₃(1/6)³(5/6)⁶*(1/6) = 84*(5)⁶/6¹⁰

Question no. 30

A fair coin is tossed 100 times. The probability of getting tails an odd number of times is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_one

Solution :
P(getting tails on odd number of times) = P(r=1) + P(r=3)... + P(r=99)
= ¹⁰⁰C₁(1/2)(1/2)⁹⁹ + ¹⁰⁰C₃(1/2)³(1/2)⁹⁷....... + ¹⁰⁰C₉₉(1/2)⁹⁹(1/2)¹
(1/2)¹⁰⁰[¹⁰⁰C₁ + ¹⁰⁰C₃+......+ ¹⁰⁰C₉₉ ]
= (1/2)¹⁰⁰2^100-1 = 1/2

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