What is the probability of rolling three fair dice and having all thre

Question

What is the probability of rolling three fair dice and having all three dice show the same odd number?

A. 71/72
B. 1/36
C. 1/72
D. 1/108
E. 1/216

quantumliner · Answer

Total number of possibilities = 6*6*6 = 216
Number of ways same odd number shows up in all three dice = 3 {(1,1),(3,3),(5,5)}
Probability of rolling three fair dice and having all three dice show the same odd number = 3/216

= 1/72

Answer is C. 1/72

JeffTargetTestPrep · Answer

The probability of the first die’s showing an odd number is 1/2. The probability of the second die’s showing the same number is 1/6. The probability of the third (or last) die’s showing the same number is also 1/6.

Thus, the overall probability is 1/2 x 1/6 x 1/6 = 1/72.

Answer: C

EgmatQuantExpert · Answer

Solution

• We need to get an odd number on each of three dice and all of them must be same.

• The odd numbers in a dice are  1,3 and 5 .

o Thus, our favourable cases are  (1,1,1) (3,3,3) ad (5,5,5) 
 
• Total cases i.e the the number of ways in which we can get numbers on 3 dice  = 6 * 6 * 6

o Hence Probability  = \frac{(Favorable cases)}{(Total Cases)} = \frac{3}{(6*6*6)} = \frac{1}{72} 
 
• Hence, the correct answer is   Option C

Thanks,
Saquib
Quant Expert
e-GMAT

Aiming to score Q50 or higher in GMAT Quant ? Attend this webinar on   2nd April   to learn a structured approach to  solve 700+ Number Properties question in less than 2 minutes .  Register

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FB2017 · Answer

Two ways to solve it
1st way:
No of favorable cases=3  
Total outcomes=6*6*6=216
Probability=No of favorable cases/Total outcomes=3/216=1/72

2nd way:
The probability of the first die’s showing an odd number is 1/2. The probability of the second die’s showing the same number is 1/6. The probability of the third (or last) die’s showing the same number is also 1/6.
Thus, the overall probability is 1/2 x 1/6 x 1/6 = 1/72.
Option C.

BrushMyQuant · Answer

We need to find What is the probability of rolling three fair dice and having all three dice show the same odd number?

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

Out of 1 to 6 there are only three odd numbers 1, 3, 5

Cases in which all three dice will show same odd number will be
(1,1,1), (3,3,3), (5,5,5) => 3 cases

=>  P(Three dice will show same odd number)  =  \frac{3}{216}  =  \frac{1}{72}

So,  Answer will be C 
Hope it helps!

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

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

bumpbot · Answer

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