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

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

Choice b is correct answer.

1/12

Step-by-step 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.