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by [email protected] » Wed Nov 01, 2017 12:28 pm
Hi swerve,

We're asked for the probability of rolling three fair 6-sided dice and having all three dice show an ODD number.

The six numbers on a die are 1, 2, 3, 4, 5 and 6 - half of those numbers are odd and half are even.

The probability of rolling an ODD number on each die is 3/6 = 1/2
The probability of rolling ODD numbers on ALL 3 dice is (1/2)(1/2)(1/2) = 1/8

Final Answer: B

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by Brent@GMATPrepNow » Thu Nov 02, 2017 8:25 am
swerve wrote:What is the probability of rolling three fair dice and having all three dice show an odd number?

A. 5/8
B. 1/8
C. 1/36
D. 1/108
E. 1/216
Each die has 6 possible outcomes (1, 2, 3, 4, 5, 6)
Of those outcomes, 3 are ODD (1, 3, 5)
So, when we role ONE die, P(ODD) = 3/6 = 1/2

P(all 3 dice are odd) = P(1st roll is odd AND 2nd roll is odd AND 3rd roll is odd)
= P(1st roll is odd) x P(2nd roll is odd) x P(3rd roll is odd)
= 1/2 x 1/2 x 1/2
= 1/8

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by Scott@TargetTestPrep » Fri Nov 01, 2019 6:54 pm
swerve wrote:What is the probability of rolling three fair dice and having all three dice show an odd number?

A. 5/8
B. 1/8
C. 1/36
D. 1/108
E. 1/216

The OA is B.

I need help to solve this PS question. Please, can any expert assist me with it? Thanks.
Since the probability of a die showing an odd number is ½, the probability of all three dice showing an odd number is 1/2 x 1/2 x 1/2 = 1/8.

Answer: B

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