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What is the probability of rolling three fair dice and having at least

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What is the probability of rolling three fair dice and having at least one of the three dice show an even number?

A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216
[Reveal] Spoiler: OA

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Re: What is the probability of rolling three fair dice and having at least [#permalink]

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We have to (easier this way) calculate first the prob to have none even number (NEN).
We have a probability of 1/2 to have an even number (2,4,6)

=> Prob(NEN) = 1/2 * 1/2 * 1/2 = 1/8

=> Prob(at least one even number) = 1- Prob(NEN) = 1-1/8 = 7/8

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Bunuel wrote:
What is the probability of rolling three fair dice and having at least one of the three dice show an even number?

A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216


We want P(select at least 1 even number)
When it comes to probability questions involving "at least," it's usually best to try using the complement.

That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 even number) = 1 - P(not getting at least 1 even number)
What does it mean to not get at least 1 even number?
It means getting zero even numbers.
So, we can write: P(getting at least 1 even number) = 1 - P(getting zero even numbers)

P(getting zero even numbers) = P(1st coin is odd AND 2nd coin is odd AND 3rd coin is odd )
= P(1st coin is odd) x P(2nd coin is odd) x P(3rd coin is odd )
= 1/2 x 1/2 x 1/2
= 1/8

So, P(getting at least 1 even number) = 1 - 1/8
= 7/8

Answer: B
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Re: What is the probability of rolling three fair dice and having at least [#permalink]

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New post 03 Feb 2017, 12:47
Bunuel wrote:
What is the probability of rolling three fair dice and having at least one of the three dice show an even number?

A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216


We can look at this problem in terms of only 2 possible events. Either at least one die shows an even number or no dice show an even number. This means:
P(at least one die showing an even number) + P(no dice showing an even number) = 1

P(at least one die showing an even number) = 1 - P(no dice showing an even number)

Thus, if we can determine the probability of no dice showing an even number, we’ll quickly be able to calculate the probability that at least one die shows an even number.

There are 3 even and 3 odd numbers on each die. Thus, the probability of not showing an even number on all three dice is 1/2 x 1/2 x 1/2 = 1/8.

Thus, the probability of showing an even number on at least one die is 1 = 1/8 = 7/8.

Answer: B
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Re: What is the probability of rolling three fair dice and having at least [#permalink]

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New post 03 Aug 2017, 02:59
P (at least one even) = 1 - P(all odd) = 1[Universe]−1/2∗1/2∗1/2=1−1/8=7/8. [Probability of all odd'd on rolling 3 dice is 1/2∗1/2∗1/2=1/8]
Option B.

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Re: What is the probability of rolling three fair dice and having at least   [#permalink] 03 Aug 2017, 02:59
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