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

Question

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

BrentGMATPrepNow · Accepted Answer

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

LHC8717 · Answer

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

ScottTargetTestPrep · Answer

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

FB2017 · Answer

P (at least one even) = 1 - P(all odd) = 1 −1/2∗1/2∗1/2=1−1/8=7/8.  
Option B.

ScottTargetTestPrep · Answer

Since P(at least one even number in 3 rolls) = 1 - P(no even numbers in 3 rolls)

and P(no even numbers in 3 rolls) = (1/2)^3 = 1/8,

then P(at least one even number in 3 rolls) = 1 - 1/8 = 7/8.

Answer: B

MathRevolution · Answer

The probability of a die showing an odd number is  \frac{1}{2} .

All three roll shows odd number:  \frac{1}{2}  *  \frac{1}{2}  *  \frac{1}{2}

Atleast 1 even: 1 - all odd = 1 -  \frac{1}{2}  =  \frac{7}{8}

Answer B

BrushMyQuant · Answer

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

As we are rolling three dice => Number of cases =  6^3  = 216

Let's solve the problem using two methods:

Method 1:

Now there are 8 outcomes possible
(Odd, Odd, Odd), (Odd, Odd, Even), (Odd, Even Odd), (Odd, Even, Even), (Even, Odd, Odd), (Even, Odd, Even), (Even, Even Odd), (Even, Even, Even) and there is an equal chance of each of them happening

=>  P(At least one number is Even)  =  \frac{7}{8}  (all cases except Odd, Odd, Odd)

Method 2:

Out of the 216 comes lets eliminate all options in which all three outcomes are odd
All three can be odd in 3*3*3 (=27 ways), as in each roll we can get any number out of 1, 3, and 5 => 3 choices in each roll

=>  P(At least one number is Even)  =  \frac{216 - 27}{216}  =  \frac{189}{216}  =  \frac{7}{8}

So,  Answer will be B 
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

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