Probability Test

Result

Probability Test - 7

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Question 1
1 / -0

Three identical dice are rolled. The probability that the same number will appear on each of them is

A
1/6

B
1/18

C
3/28

D
1/36

Solution

Since throwing a single die three times is equivalent to throw three dice at a time

∴ Sample space = {(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6), .......}

Here, n(5) = 6³

∴ Required Probability =6/6³= 1/6²= 1/36
Question 2
1 / -0

One ticket is selected at random from 100 tickets numbered 00, 01, 02…, 99. Suppose S and T are the sum and product of the digits of the number on the ticket, then the probability of getting S = 7 and T = 0 is

A
19/100

B
1/50

C
1/4

D
2/19

Solution

Given 100 tickets numbered 00, 01, 02…, 99. , S and T are the sum and product of the digits of the number on the ticket. here the number of elements in the sample space = 100

Now,S=7 i.e. sum of the digits is 7,S={07,70,16,61,25,52,34,43}, and T=0 i.e. product of the digits = 0 ,T={ 00,01,02,03,04,05,06,07,08,09,10,20,30,40,50,60,70,80,90}

T=0 and S=7 ⇒ T ∩ S ={07,70} ⇒ n(T ∩ S) = 2

Hence P(T=0 and S=7)= P( T ∩ S) = 2/100 = 1/50
Question 3
1 / -0

A coin tail on both sides is tossed twice. The probability of getting ‘a head’ is

A
0

B
1

C
3/4

D
1/2

Solution

Since a coin tail has both faces as tail.so getting head is an impossible event .so its probability is zero
Question 4
1 / -0

Mean number of ‘sixes’ in four throws of a fair dic is

A
6/4

B
1/4

C
1

D
4/6

Solution

Let X be a random variable denoting the no. of sixes. Then X follws B(4, P)

Here, p Denotes the probability of getting a six in a single throw of a die

∴Required probability = nP

= 4 × 1/6 = 4/6
Question 5
1 / -0

From a deck of 52 cards, the probability of drawing a court card is

A
1/4

B
1/13

C
3/13

D
4/13

Solution

Since there are 12 court cards in a pack of 52 cards. Therefore,

Required probability = 12/52=3/13