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

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

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

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

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.

Choose out of 3 attempts which one will have 3 i.e in 3c1 ways
Multiply probabilities of getting 3 i.e 1/6 and a number other than 3 i.e 5/6
3C1 * 1/6*5/6*5/6 =\(\frac{ 25}{71}\)

Option (C)

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!

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

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