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Probability - 3 rolls of a number cube

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Probability - 3 rolls of a number cube

by Valentino99 » Sun Aug 26, 2012 6:30 am
Hello, its a simple problem but I am not able to spot my fault. Please assist.

Q) what is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be 6?

Solution:
Let A be the event that the number cube rolls a 6.
P(A) = 1/6
P(not A) = 5/6

As it is asked, what is probability of at least one of the rolls to be 6,
we can quickly calculate the probability of no rolls to be 6, which is equal to (5/6)*(5/6)*(5/6) = 125/216.
Hence, probability of at least one of the rolls to be 6 = 1 - 125/216 = 91/216

My doubt:
If I think of calculating the probability in direct way, please tell me where I am making mistake.

We are asked : prob. of at least one of the rolls be a 6 == it means, one roll could be a 6 OR two rolls could be a 6 OR all 3 rolls could be a 6

therefore, ans = P(one roll be a 6) + P(two rolls be a 6) + P(all 3 rolls be a 6)
ans = ((1/6)*(5/6)*(5/6)) + ((1/6)*(1/6)*(5/6)) + ((1/6)*(1/6)*(1/6))
ans = 31/216

Please help me correct my discrepancy between getting the answer as 91/216 and 31/216.
Thanks!
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by Brent@GMATPrepNow » Sun Aug 26, 2012 7:02 am
Valentino99 wrote:Hello, its a simple problem but I am not able to spot my fault. Please assist.

Q) what is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be 6?


We are asked : prob. of at least one of the rolls be a 6 == it means, one roll could be a 6 OR two rolls could be a 6 OR all 3 rolls could be a 6

therefore, ans = P(one roll be a 6) + P(two rolls be a 6) + P(all 3 rolls be a 6)
ans = ((1/6)*(5/6)*(5/6)) + ((1/6)*(1/6)*(5/6)) + ((1/6)*(1/6)*(1/6))
ans = 31/216

Please help me correct my discrepancy between getting the answer as 91/216 and 31/216.
Thanks!
The part in green (above) is perfect. However, the part in blue, is missing some pieces.

Let's focus on: P(one roll is a 6)
You wrote that this probability = (1/6)(5/6)(5/6)
This is not correct. This solution considers only the outcome of {6, not 6, not 6}
You also need to consider the outcomes {not 6, 6, not 6} and {not 6, not 6, 6}

So, P(one roll is a 6) = (1/6)(5/6)(5/6) + (5/6)(1/6)(5/6) + (5/6)(5/6)(1/6)
= 75/216

Similarly, for P(two rolls are 6), your solution of (1/6)(1/6)(5/6) considers only one possible outcome {6, 6, not 6}. You also need to consider {6, not 6, 6} and {not 6, 6, 6}

So, P(two rolls are 6) = (1/6)(1/6)(5/6) + (1/6)(5/6)(1/6) + (5/6)(1/6)(1/6)
= 15/216

Your solution for P(all 3 rolls are 6) is perfect.
P(all 3 rolls are 6) = (1/6)(1/6)(1/6)
= 1/216

So, put everything together to get:
P(at least one 6) = 75/216 + 15/216 + 1/216
= 91/216

I hope that helps.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Valentino99 » Sun Aug 26, 2012 7:09 am
:) Yes it does. Thank you!

I gave a calm thought and figured it out too but at the time of the post, I was not able to figure it out..!

Thanks for your reply!!
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