What is the probability of rolling two fair dice and having neither

Question

What is the probability of rolling two fair dice and having neither die show a prime number?

A. 35/36
B. 3/4
C. 1/2
D. 1/4
E. 1/36

vitaliyGMAT · Answer

There are 3 prime numbers shown on the die: 2, 3 and 5.

The probability of showing a prime number on a single die is 1/2, hence the probability of not showing a prime number is also 1/2.

Events are independent: 1/2*1/2 = 1/4

Answer D

shashank1tripathi · Answer

Total number of cases when two dice show prime number is 9.(Multiple sets for values 2,3 or 5 on either of two dice)
Total possible outcomes when two dice are rolled = 6*6 = 36.
Probability to get prime numbers on both the dice = 9/36 = 1/4
Therefore, probability of none of two dice shows prime number = 1 - (1/4) = 3/4
Ans. B

vitaliyGMAT · Answer

Let's use your logic to find the probability that dice will show a nimber, which is not prime but will start from non prime this time.

Total number of cases when two dice show a number which is not prime is 9, total possible outcomes 36.

Probability of getting a non prime number on both dice simultaniously 9/36 = 1/4.

Probability of getting pime number hence is 1-1/4 = 3/4

Now, because our numbers on a dice are either prime or non prime then their sum will compose full set of possible outcomes.

3/4 + 3/4 = 6/4 which is greater than 1 and that's impossible, because we know that probability can not be greater than 1.

The problem is when you take the diffrence 1- 1/4 yo are subtracting probabiliy that two dice will show a prime number from a total set, which includes the outcomes:

both dice show a nimber which is not prime + dice #1 shows a number which is not pime + dice #2 shows a number which is not prime .

That is you'll get probability that  AT LEAST ONE  dice will show the number which is not pime and that probabiliy is bigger than the pobability of both dice shonwing non prime number simultaniously. That's why we are getting sum >1.

Now let's compose the whole set of outcomes:

dice#1(prime) dice#2(prime) + dive#1(non prime) dice#2(non pime) + dice#1(prime) dice#2(non prime) + dice#1(nonprime) dice#2(prime) =

\frac{3}{6}*\frac{3}{6} + \frac{3}{6}*\frac{3}{6} + \frac{3}{6}*\frac{3}{6} + \frac{3}{6}*\frac{3}{6} = \frac{36}{36} = 1

shashank1tripathi · Answer

@vitalityGMAT - Thanks for correcting me. I got the point that by subtracting 3/4 from 1, I included the set of values for which one dice shows primary number.

BrushMyQuant · Answer

We need to find What is the probability of rolling two fair dice and having neither die show a prime number?

As we are rolling two dice => Number of cases =  6^2  = 36

We know that out of the numbers from 1 to 6 we have only three prime numbers and they are 2, 3 and 5
( Watch this video  if you are not aware of  Prime Numbers )

And neither of the die should show a prime number
=> For each die we have only 3 numbers possible, 1, 4 and 6
=> We have 3 possibilities for the first die and for each of these 3 possibilities we have 3 possibilities for the second die
=> 3*3 = 9 possibilities

=> Probability that neither die show a prime number =  \frac{9}{36}  =  \frac{1}{4}

So,  Answer will be D 
Hope it helps!

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

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

What is the probability of rolling two fair dice and having neither

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