In a game, one player throws two fair, six-sided die at the

Probably the easiest approach would be to find the probability of the opposite event and subtract it from 1:

P(win) = 1- P(not win) = 1 - 4/6*4/6 = 5/9.

Answer: C.
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We have 2 good (read five and one) possibilities (\(G\)) on 6 faces G=2/6 and 4 bad possibilities (\(B\)) on 6 faces B=4/6
The winning combinations are the ones with at least a \(G\) in it so:
\(G,B\)
\(B,G\)
\(G,G\)

\(G,B\) and \(B,G\) have the same probability \(\frac{2}{6}*\frac{4}{6}=\frac{2}{9}\) each
\(G,G\) has a probability of \(\frac{2}{6}*\frac{2}{6}=\frac{1}{9}\)
Sum them up \(\frac{2}{9}+\frac{2}{9}+\frac{1}{9}=\frac{5}{9}\)

So, the player wins if he/she rolls AT LEAST one 5 or 1.

When it comes to probability questions involving "at least," it's best to try using the complement.

That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= 5/9

Answer: C

Cheers,
Brent

Option C.

the number of cases in which he can lose the game are when both the faces have neither of 5 or 1 or both. so the possible combinations are (2,2),(2,3),(2,4) (2,6) and 12 more with 3,4,6.

probability of loss = # loss cases/# total no of cases
= 16/36 or 4/9

hence probability of win = 1-p(loss). = 1-(4/9) = 5/9

WHy in both winning combination we are calculating for GG only once .
May be on first die 5 and second die one or on first die one and second die 5...These can combinations can also occur na? ..WHy we are not considering this scenario?

Hi skamal7,

Consider the following example that will explain better than any theoretical information.
You say that G,G should be counted twice, so the possible combinations are:

G,G=1/9
G,G=1/9
B,G=2/9
G,B=2/9
B,B=4/9
[ also if your method is correct B,B should be counted twice =4/9 ]

don't you see anything odd? The sum of the probability of each case is greater than 1! \(\frac{1+1+2+2+4}{9}=\frac{10}{9}\)
[ if you count B,B twice it becomes \(\frac{14}{9}\) ]

Why does this happen?Let's look at the theory now
The formula to solve this problem is \((nCk)p^k*q^{(n-k)}\) where p=1/3 and q=2/3 and N are the dies and K are the good outcomes:

Case two good \((2C2)(\frac{1}{3})^2(\frac{2}{3})^0=\frac{1}{9}\)
Case one good one bad \((2C1)(\frac{1}{3})^1(\frac{2}{3})^1=\frac{4}{9}\)
Case two bad \((2C0)(\frac{1}{3})^0(\frac{2}{3})^2=\frac{4}{9}\)

Tot sum = \(\frac{1+4+4}{9}=1\)

Hope it's clear now, let me know

I think the question should be re-worded. 'At least a five' sounds like >= 5. Therefore, my result was 1-(1/2*1/2) = 3/4

Hi All,

This question can be solved with "brute force." Since you're rolling 2 dice, there aren't that many possible outcomes (just 36 in total), so you COULD just write them all down:

We're looking for the number of outcomes that include AT LEAST a 1 or a 5.

1,1
1,2
1,3
1,4
1,5
1,6

2,1
2,5

3,1
3,5

4,1
4,5

5,1
5,2
5,3
5,4
5,5
5,6

6,1
6,5

Total possibilities = 20

Probability of rolling at least a 1 or a 5 on two dice: 20/36 = 5/9

Final Answer:

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

hi experts!

when I was solving this question, I merely added the probability of the two dices rolling a '5' or a '1' each; 2/6 + 2/6 = 2/3 since both are independent events.

which scenario did I overcount and when should I be solving the opposite events then subtracting it from 1? I've been solving over 50 probability questions and I'm still not getting a hang of it.

Hi whitehalo,

You 'double-counted' scenarios in which you roll a 1 or a 5 on BOTH dice.

1,1
1,5
5,1
5,5

Each of these options should be counted just ONCE, but your math counts them twice (thus, incorrectly raising your answer to a higher probability).

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

This one is not an Official gmat question; It would be great, if those math experts writing such low-end stems firstly could get trained in SC. I lost much time trying to solve this one before I finally realized that the thing asked is actually different from what is meant. There is a parallelism issue with ‘at least’ here. Ideally, ‘at least’ is a redundant phrase and instead the stem should read as ‘if the player receives a five or a one on either side...’ This writing will include all the cases meant by the original stem and exclude those without 5 or 1 on either side.

at least based
so not getting 1 or 5 = 4/6
for two dice = 4/6 * 4/6 ; 4/9
1-4/9 = 5/9
IMO C

hey buddy can you let me know why this does not work
prob(1 or 5 in first die) = 2/6=1/3
prob(1 or 5 in second die) = 1/3
So, required prob = 1/9

Hi AnirudhaS,

Your calculation is for the probability that BOTH dice show a '1' or a '5' - but the question asks for a '1' or '5' on EITHER of the dice. This means that there are 3 possible ways to win:

1) The 1st die matches and the 2nd does not: (1/3)(2/3) = 2/9
2) The 2nd die matches and the 1st does not: (1/3)(2/3) = 2/9
3) Both the 1st and 2nd die match: (1/3)(1/3) = 1/9

Total probability of getting at least one '1' or '5' = 2/9 + 2/9 + 1/9 = 5/9

You could also have solved this question by calculating the probability of NOT winning and subtracting that from 1. That would be...

1 - (2/3)(2/3) =
1 - 4/9 =
5/9

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

Given that In a game, one player throws two fair, six-sided die at the same time. If the player receives at least a five or a one on either die, that player wins. And we need to find What is the probability that a player wins after playing the game once?

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

To win the player needs to get at least a five or a one on either die
=> Player can get 5 or 1 or 1, 5 on at least one die to win the game

So, possible cases are

Case 1: First roll get any number out of 1 or 5 in 2 ways
Second die get any number on the other die in 6 ways
=> 2*6 = 12 ways

Case 2: First roll get any number out of the six numbers except 1 or 5 as they are already considered in Case 1 => 4 ways
Second die get any number out of 1 or 5 in 2 ways
=> 4*2 = 8 ways

=> Total cases = 12 + 8 = 20

=> Probability that a player wins after playing the game once = \(\frac{20}{36}\) = \(\frac{5}{9}\)

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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I agree with ShukhratJon. This particular question is not properly worded.

Yes, the question stem should read "If the player receives a five or a one on at least one die..."
I thought it is saying "at least a 5" or "a one" and hence proceeded to consider 5, 6 and 1. Though even this interpretation wasn't satisfactory because I wondered if "at least a 5 or a 1" meant "at least a 5" or "at least a 1" which would make no sense since every throw of a die will give at least a 1.
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