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

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

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

98%

step-by-step explanation: