Question

You roll a number cube twice. What is the probability of rolling a 2 first and then rolling an odd number? 1/36,1/12,1/9,1/4

Answer 1

A = event of rolling a 2 on the first roll

P(A) = 1/6 because there is only one "2" out of six labels total on the number cube.

B = event of rolling an odd number on the second roll

P(B) = 3/6 = 1/2 since there are three odd numbers {1,3,5} out of six total

The events A and B are independent, so we can multiply the probabilities

P(A and B) = probability of both events happening simultaenously

P(A and B) = P(A)*P(B)

P(A and B) = (1/6)*(1/2)

P(A and B) = (1*1)/(6*2)

P(A and B) = 1/12

Final Answer: 1/12

Note: The fraction 1/12 is approximately equal to 0.0833

Answer 2

Choice b is correct answer.

1/12

Explanation:

We have to find the probability of happening of two events.

From question statement,

total outcomes = 6

Let

A = rolling of a 2 (favourable outcomes = 1)

B = rolling of an odd number(1,3,5) (favourable outcomes = 3)

Probability is ratio of favourable outcomes to total outcomes.

P(A) = 1/6

P(B) = 3/6 = 1/2

The probability of two events occurring smiltaneously,

P(A and B ) = P(A) P(B)

Putting the values of P(A) and P(B) . we have

P(A and B ) = (1/6)(1/2)

P(A and B ) = 1/12 which is the answer.