If a fair 6-sided die is rolled three times, what is the probability

Question

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6

AaronPond · Accepted Answer

Here is the full answer, from the ground up. To understand the mathematics behind this question we need to understand three basic probability rules:

(1)    A probability is calculated by dividing the number of options when the event occurs by the total number of possibilities.

(2)    When calculating the probability of multiple events that must ALL occur to meet a certain condition, we must    multiply    the separate probabilities of each event together. Thus, the probability of all the events  A ,  B , and  C  occurring would be:  P_{combined} = (P_A)*(P_B)*(P_C)

(3)    When calculating the probability of a series of mutually-exclusive events, where ANY event occurring would be sufficient, we must    add    the separate probabilities of each event together. Thus, the probability of any of the events  A ,  B , and  C  occurring would be:  P_{combined} = (P_A)+(P_B)+(P_C)

Let's leverage these three basic rules to solve this problem.

First, the probability of rolling a single value (in this case 3) on a fair 6-sided die would be one out of six (or  \frac{1}{6} ).

The probability of NOT rolling that number would be five out of six (or  \frac{5}{6} ).

So, if you need to roll a single value on one die, while simultaneously not rolling that value on the other two dice, then we invoke rule #2 from above (i.e., we multiply the probabilities.) Thus,

P=(\frac{1}{6})*(\frac{5}{6})*(\frac{5}{6})=\frac{25}{216}

Notice that this is answer choice  A . However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work. There are three possibilities:

3, N, N 
 N, 3, N 
 N, N, 3

The probability of each of these orders is the same ( \frac{25}{216} ), but each is separate and independent. We now need to invoke rule #3 from above and add the separate probabilities together:

P_{Total} = \frac{25}{216} + \frac{25}{216} + \frac{25}{216} = \frac{3*25}{216}

Recognizing common factors in the top and bottom of this fraction allow us to get to the answer without any messy math.  216 = 6*6*6 = 2*3*6*6 . Thus,

\frac{3*25}{216}=\frac{3*25}{2*3*6*6}=\frac{25}{2*6*6}=\frac{25}{72}

The answer is C.

KarishmaB · Answer

Total ways in which a 6-sided die can be rolled three times = 6*6*6 = 216

To get exactly one 3, there are three ways:
A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways.
The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75

Required Probability = 75/216 = 25/72

Answer (C)

niel1989 · Answer

Probability if getting 3 on a die is: P(3)=1/6 
Probability of not getting 3 on a die is nP(3)=5/6

Let the three dies be denoted by P1,P2 and P3

Probability of getting 3 on the first die and not getting 3 on the other two dies is given by:-
=P1(3)*nP(3)*nP(3)
=1/6*5/6*5/6
=25/216

As there are 3 dies,so similarly for the rest of the two dies probability will be 25/216

Summing up all the 3 cases we get:
=3*25/216
=25/72

MathRevolution · Answer

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6

Let O be represented as 3 being rolled, and X be represented as no 3 being rolled.

Then the case of exactly one 3 being rolled can be represented as (O, X, X), (X, O, X), (X, X, O). 
The probability of (O, X, X) is (1/6)*(5/6)*(5/6), that of (X, O, X) is (5/6)*(1/6)*(5/6) and that of (X, X, O) is (5/6)*(5/6)*(1/6).
So the probability is (1/6)*(5/6)*(5/6) + (5/6)*(1/6)*(5/6) + (5/6)*(5/6)*(1/6) = 3*(1/6)*(5/6)*(5/6) = 25/72.

The answer is (C).

ydmuley · Answer

Probability of getting 3  = \frac{1}{6}

Probability of not getting 3  = \frac{5}{6}

Probability of getting three on the first roll  = \frac{1}{6} * \frac{5}{6} * \frac{5}{6}

Probability of getting three on the first roll  = \frac{25}{216}

As we can get 3 on either second or the third time we will multiply the above probability by 3

= \frac{25}{216} * 3

= \frac{25}{72}

Hence, Answer is C

EMPOWERgmatRichC · Answer

Hi All,

You can approach the math in this question in a couple of different ways, so you have to think about what would be the easiest way for YOU to organize your work.

The questions tells us to roll a 6-sided dice 3 times. We're asked for the probability of rolling EXACTLY one '3' on those three rolls.

Here's a way to break the calculation down into 3 smaller calculations:

(first roll is 3)(second roll is NOT 3)(third roll is NOT 3) = (1/6)(5/6)(5/6) = 25/216

(first roll is NOT 3)(second roll is 3)(third roll is NOT 3) = (5/6)(1/6)(5/6) = 25/216

(first roll is NOT 3)(second roll is NOT 3)(third roll is 3) = (5/6)(5/6)(1/6) = 25/216

Total = 3(25/216) = 75/216 = 25/72

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

saumya2805 · Answer

"A" die is rolled 3 times.. There aren't 3 dies... which is why I can't seem to understand why not 25/216

saumya2805 · Answer

You mention the below:
"Notice that this is answer choice A. However, this is a deliberate trap answer.   With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.)   Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work."

Plz refer the bold colored portion..
Is it understood by default, that the order always matters? Even if it isn't mentioned explicitly in the question?
This is exactly what I'm always confused about in probability related questions.

AaronPond · Answer

Thank you, Saumya2805, for the question. Yes, when calculating the probability of multiple situations, the order always matters. This isn't just the "default", this is how the mathematics of probability work. Even if the events are simultaneous, think about the solution one event at a time. Each different arrangement has a separate probability.

KarishmaB · Answer

How do you get 216?
You say the FIRST roll could be 1/2/3/4/5/6 i.e. 6 ways
The SECOND roll could be 1/2/3/4/5/6 i.e. 6 ways
The THIRD roll could be 1/2/3/4/5/6 i.e. 6 ways

Hence rolling a 3 on FIRST roll and any of the other 5 numbers of other two rolls is different from rolling a 3 on SECOND roll and any of the other 5 numbers on other two rolls because we have counted these cases apart {3, 1, 6}, {1, 3, 6}. That is how we get 75 and not just 25.

This is same as saying - I have 3 dice of different colours - Red, Yellow and Blue - and I roll them together 
The Red could have outcome in 6 ways, Yellow in 6 and Blue in 6 so 216 ways

Same as case above.

How about saying I have 3 identical dice and I roll them together. How many outcomes do I have in that case?
Is (1, 2, 3) different from (3, 1, 2)? No. Then I do not have 216 outcomes in this case. In this case, there is no "order".

ScottTargetTestPrep · Answer

P(Exactly one three) = 1/6 x 5/6 x 5/6 = 25/216

The number of ways to select one 3 out of three rolls is YNN (or NYN or NNY) = 3!/2! = 3 ways.

So, the total probability is 3 x 25/216 = 75/216 = 25/72.

Answer: C

CrackverbalGMAT · Answer

Solution:

Total sample space here = 6^3 = 216

There are two cases two consider for a favorable outcome:-

Case 1 :- A 3 on the first roll of the die and a different number other than 3 on the next 2 rolls and this is possible in

1 way * 5 ways * 5 ways =25 OR

Case 2:- A 3 on the second roll and a different number other than 3 on the other 2 rolls and this is possible in

5 ways * 1 way * 5 ways = 25 OR

Case 3:- A 3 on the third roll and a different number other than 3 on the other 2 rolls and this is possible in

5 ways * 1 way * 5 ways = 25

The total number of ways possible (Favorable cases) = 25 + 25 +25 =75

=>Probability = 75/216  (option c)
 
  Devmitra Sen
GMAT SME

Chitra657 · Answer

I got this question today in my mock and got it wrong. 
I did correctly till the 1/6*5/6*5/6 step
but then I multiplied the whole thing by 3! thinking that all those three numbers would be different, i.e., one would be three, other two can be any two numbers(not necessarily the same), so I did not divide 3! by 2!. 
Can you pls clear my confusion?  VeritasKarishma   Bunuel   ScottTargetTestPrep

EMPOWERgmatRichC · Answer

Hi Chitra657,

To start, it's worth noting that most GMAT questions are written so that they can be approached in more than one way (so if you find one approach to be unclear or too complicated, then there is likely another approach that you would find faster or easier). With this question, it does not matter which of the three rolls is a '3', so we're ultimately looking for all of the COMBINATIONS of three rolls that 'fit' what we're looking for (and the overall probability of hitting one of those options). Thus, the last part of this calculation would be 3c1 (meaning 1 of the 3 rolls is a '3') OR 3c2, (meaning 2 of the 3 rolls are NOT '3s') - since those two calculations lead to the same result.
 
GMAT assassins aren't born, they're made,
Rich

KarishmaB · Answer

Chitra657  -

Note what this step 1/6*5/6*5/6 gives you:

3-4-2 or 3-6-1 or 3-2-4 or 3-6-1 etc.
The second and third roll can be anything other than 3. So we will include all cases such as 3-4-2 as well as 3-2-4.

If you multiply now by 3!, you are again re-arranging these numbers to get a lot of duplicate cases. 
3-4-2 will give 4-3-2 and 4-2-3 but also 3-2-4 etc. But these are already accounted for when we do 1/6*5/6*5/6.

So all we need to do is multiply by 3 so that
3-4-2 gives 4-3-2 and 4-2-3. Those are the only extra cases we need to include. 
Then our 3-2-4 case will give us 2-3-4 and 2-4-3.

BrushMyQuant · Answer

Given that a fair 6-sided die is rolled three times and We need to find what is the probability that exactly one 3 is rolled?

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

Now we have three places in these 3 tosses _ _ _

First of all let's find the toss in which 3 will come. This can happen by selecting 1 place out of 3 where 3 can come.
=> 3C1 = 3 ways

Now, Probability of getting a 3 in any toss =  \frac{1}{6}  (As there is one 3 out of the 6 possible outcomes)
P of not getting a 3 in any toss =  \frac{5}{6}  (As there are 5 outcomes out of 6 in which 3 doesn't come)

=> Probability that exactly one 3 is rolled = Place of that 3 * Probability of getting a 3 * P of not getting a 3 in any toss * P of not getting a 3 in any toss = 3 *  \frac{1}{6}  *  \frac{5}{6}  *  \frac{5}{6}  =  \frac{25}{72}

So,  Answer will be C 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

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

dns1357 · Answer

Hi KarishmaB,

Thanks for posting the solution. I have a very basic doubt. Should we not multiply by 3! rather than just 3. I always make mistake in such cases where we have to multiply by the no of cases and factorials. Requesting a conceptual clarity on the same.

BrushMyQuant · Answer

Here also we are doing factorial only, it is just cancelling out.

We need to find the position of when we are going to get that 3 out of the 3 possible places. 
This can be done in 3C1 ways =  \frac{3!}{1!*(3-1)!} =  \frac{3 * 2!}{1!*2!}  = 3 ways

Hope it helps!

If a fair 6-sided die is rolled three times, what is the probability

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If a fair 6-sided die is rolled three times, what is the probability

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