What is the probability that, on three rolls of a number cub

Question

Hi,

This question is actually from the Manhattan Number Properties book,

I am trying to solve this, but am getting it wrong.

What is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be a 6?

The way I am solving it:

But the right answer is  \frac{91}{216} . The book has given this example to explain the 1-x probability trick, ie. calculating the probability of not getting even one 6 and subtracting it from 1.

I was trying to solve it the normal way, to understand it.

I seem to get 91/216 if I consider WHICH ROLL is a 6 in the first two outcomes, and add their probabilities. 
eg, for Outcome 1: First roll is a 6 AND the roll 2 and Roll 3 are not OR the Second roll is a 6 and the other two are not OR the third roll is a 6 and the other two are not

This would give me  \frac{25}{216}+\frac{25}{216}+\frac{25}{216} = \frac{75}{216}  for Outcome 1 
and  \frac{5}{216}+\frac{5}{216}+\frac{5}{216} = \frac{15}{216}  for Outcome 2, Outcome 3 would be  \frac{1}{216}  for a total of  \frac{91}{216}

But I don't see why WHICH ROLL is a 6 is important for this problem.

What am I thinking wrong?

Zarrolou · Accepted Answer

What you say is correct, but not complete.

Case 1: only one 6. you here analyze one case (6,N,N) correctly, but you forget to analyze the other 2 cases : (N,6,N) and (N,N,6)
so the probability of case one is  3*\frac{25}{216}

The same happens for case 2: two 6s. Here again we have to multiply your result by 3 in order to get all the combinations: (N,6,6)(6,N,6)(6,6,N)
 3*\frac{5}{216}

The probability of the third case is correct. Now we sum our results and obtain  \frac{91}{216}

rooker · Answer

Thanks Zarrolou.

I'm having trouble understanding the concept of probability. 
Im trying to frame this question as a Combinatorics question. Please help me out.

Lets just consider Case 1, Only one 6 in three rolls.

So, for three rolls, all the possible outcomes are:

6 outcomes of first roll AND 6 of second AND 6 of third = 216

So, the right way to frame the question now would be :  What are the different ways of getting exactly one 6 in three rolls of a dice?

Using the Slots method, that would be:

1  AND  5  AND  5   OR  5  AND  1  AND  5  OR  5  AND  5  AND  1 .

That would be 75.

So, probability of getting exactly one 6 would now be,  \frac{Ways of getting desired outcome}{All possible outcomes}  =  \frac{75}{216}

I am sorry for taking this roundabout rote, but am just trying to get it clear for myself.

Zarrolou · Answer

Absolutely correct! The probability of getting one 6 and three N6s is:  \frac{1}{6}*\frac{5}{6}*\frac{5}{6}  
The number of possible outcome is which this can happen is 3. So  3*\frac{25}{216}  is the magic number

smartyman · Answer

Hi there,

This is my way of approach:
The probability of not getting a 6 in 3 throws is (5/6)X(5/6)x(5/6).
Therefore the probability of getting at least one 6 is 1-   = 91/216

EMPOWERgmatRichC · Answer

Hi All,

When it comes to probability questions, there are only 2 things that you can figure out:

1) What you WANT to have happen. 
2) What you DON'T WANT to have happen.

If you add those two things together, then you get the number 1.

For some probability questions, it's pretty easy to figure out the exact thing that you want to have happen. In other questions though, it's easier to figure out what you DON'T WANT, then subtract that probability from 1 (and you'll have the probability of what you DO want).

In the example that you posted, we're asked for the probability of rolling 3 dice and getting AT LEAST one 6. That means getting one 6, two 6s or three 6s would fit the definition of what we WANT. That seems like a LOT of different possibilities to keep track of. It's easier to figure out what we DON'T WANT (which would be zero 6s).

With 3 dice rolls, the probability of getting zero 6s is....

(not a 6)(not a 6)(not a 6) = (5/6)(5/6)(5/6) = 125/216

1 - 125/216 = 91/216 = the probability of getting AT LEAST ONE 6.

As you deal with additional probability questions during your studies, you have to be clear on what the question asks you to figure out, then think about which of the two approaches would be easier.

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

Yukta20 · Answer

Probability of getting at least 1 six in three rolls = 1- (no six in three rolls)

Probability of no six in 3 rolls = (5/6) * (5/6) * (5/6) = (125/216)

Thus, answer = 1- (125/216) = (91/216)

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

What is the probability that, on three rolls of a number cub

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

What is the probability that, on three rolls of a number cub

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards