Practice question on probability for NDA , Airforce and iit jee page no. 3

Probability

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Question no. 31

A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is

looks_one

looks_two

looks_3

looks_4 none of these

Option looks_3

Solution :
P(x=7) = P(x=9)

P(r=2) = ¹⁶C₂(1/2)¹⁶ = 15/2¹³

Question no. 32

A rifleman is firing at a distant target and has only 10% chance of hitting it. The least number of rounds, he must fire in order to have more than 50% chance of hitting it at least one is

looks_one 11

looks_two 9

looks_3 7

looks_4 5

Option looks_3 7

Solution :
P(hit)= 0.1
P(at least one) = P(x=1) + P(x=2)....+P(x=n) = 1-P(x=0)
P(at least one) > 0.5
1-P(x=0) > 0.5
0.5 < P(x=0)
0.5 < (0.9)ⁿ
n = 7

Question no. 33

If in a binomial distribution n =4 , P(X=0) = 16/81, Then P (X = 4) equals

looks_one 1/16

looks_two 1/81

looks_3 1/27

looks_4 1/8

Option looks_two 1/81

Solution :
P(X=0) = ⁴C₀(p)⁰(q)⁴
16/81 = (q)⁴
q = 2/3
p+q = 1
p = 1-2/3 = 1/3

Question no. 34

In A box containing 100 bulbs, 10 are detective. What is the probability that out of a sample of 5 bulbs, none is defective

looks_one

looks_two

looks_3 10^-5

looks_4

Option looks_one

Solution :
probability that out of a sample of 5 bulbs, none is defective = ⁵C₅(90/100)⁵(10/100)⁰ = ⁵C₅(9/10)⁵ = (9/10)⁵

Question no. 35

From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24 is

looks_one 6/25

looks_two 1/4

looks_3 1/6

looks_4 None of these

Option looks_4 None of these

Solution :
Number obtained on the card is divisible by 6 or 8 but not by 24 = 6,8,12,16,18,30,32,36,40,42,54,56,60,64,66,78,84,88,90
The probability that the number obtained on the card is divisible by 6 or 8 but not by 24 = 20/100 = 1/5

Question no. 36

The probability that in a year of 22nd century chosen at random, there will be 53 Sundays is

looks_one 3/28

looks_two 2/28

looks_3 7/28

looks_4 5/28

Option looks_one 6 sq. unit

very soon 6

Question no. 37

Two persons A and B take turns in throwing a pair of dice. The first person to throw 9 from both dice will be awarded the prize. If A throws first, then the probability that B wins the game is

looks_one 9/17

looks_two 8/17

looks_3 8/9

looks_4 1/9

Option looks_two

very soon

Question no. 38

A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black bells. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is

looks_one 2/15

looks_two 7/15

looks_3 8/15

looks_4 14/15

Option looks_two not cross each other

very soon

Question no. 39

If P ( A U B) = 0.8 and P(A ∩ B)=0.3, then P (A)+ P(B)=

looks_one 0.3

looks_two 0.5

looks_3 0.7

looks_4 0.9

Option looks_4 0.9

Solution :
P (A) = 1 - P ( A U B) + P(B)
P(B) = 1 - P ( A U B) + P(A)
P (A)+ P(B)= 2 -2P ( A U B) + P(A) + P(B)
P (A)+ P(B) = 2 - 1.6 + 0.5 = 0.9

Question no. 40

If A and B are two events, then P( A ∩ B) =

looks_one P(A ) P(B)

looks_two 1- P(A) - P(B)

looks_3 P(A) + P(B) - P(A ∩ B)

looks_4 P(B) - P(A ∩ B)

Option looks_4 P(B) - P(A ∩ B)

Solution :

P( A ∩ B) = P(B) - P(A ∩ B)

Question no. 41

If S is the sample space and P (A) = and S= A U B , where A and B are mutually exclusive events, then P (A) =

looks_one 1/4

looks_two 1/2

looks_3 3/4

looks_4 3/8

Option looks_one 1/4

Solution:
P(A) = P(B)/3
S = A ∪ B
A and B are mutually exclusive
So , P(A∪B) = P(A) + P(B)
1 = P(A) + 3P(A)
P(A) = 1/4

Question no. 42

A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the same colour is

looks_one

looks_two

looks_3

looks_4

Option looks_4

Solution :
P(two brown) = (5✕4)/(9✕8)
P(two white) = (4✕3)/(9✕8)
probability that these are of the same colour = (5✕4)/(9✕8) + (4✕3)/(9✕8) = 4/9

Question no. 43

A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, the probability that it is rusted or is a nail is

looks_one3/16

looks_two 5/16

looks_3 11/16

looks_4 14/16

Option looks_3 11/16

Soluion :
P(Rusted or nail) = P(rusted) + P(nail) -P(Rusted and nail)
P(Rusted or nail) = 8/16 + 6/16 - 3/16 = 11/16

Question no. 44

A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is

looks_one 64/64

looks_two 49/64

looks_3 40/64

looks_4 24/64

Option looks_one 64/64

Solution :
Probability (Good) = 10/16
Probability (defect) = 6/16
probability that it is either good or has a defect = 10/16 + 6/16 = 16/16 = 1

Question no. 45

Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floor is

looks_one

looks_two

looks_3

looks_4

Option looks_one

Solution :
Probability of all 5 persons leaving at different floor = (7✕6✕5✕4✕3)/(7✕7✕7✕7✕7) = ⁷P₅/7⁵

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