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

Question

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. What is the probability that a player wins after playing the game once?

A. 1/3
B. 4/9
C. 5/9
D. 2/3
E. 3/4

A. 1/3
B. 4/9
C. 5/9
D. 2/3
E. 3/4

Bunuel · Accepted Answer

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.

Zarrolou · Answer

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}

BrentGMATPrepNow · Answer

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 -    x   P(no 5 or 1 on 2nd die)] 
= 1 -    x    4/6]
= 1 -  
= 20/36
= 5/9

Answer: C

Cheers,
Brent

mdbharadwaj · Answer

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

skamal7 · Answer

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?

Zarrolou · Answer

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

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}  
 \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

EMPOWERgmatRichC · Answer

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:  C

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

whitehalo · Answer

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.

EMPOWERgmatRichC · Answer

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

Archit3110 · Answer

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

AnirudhaS · Answer

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

EMPOWERgmatRichC · Answer

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

BrushMyQuant · Answer

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

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

houston1980 · Answer

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

KarishmaB · Answer

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.

billlyjoehbs · Answer

KarishmaB   Bunuel

What is the probability that a player wins  after  playing the game once?"

Is there anything wrong in this sentence? I got confused with the above statement by interpreting "wins after playing the game once" as that he failed to win in the first attempt and that i need to find the probability of winning in second or third or fourth.... upto infinity. So I basically reworded the question as meaning the the chance of him winning in any other attempt other that 1st and approached it as follows.

total probability = 1
Total probability = P(winning in 1st attempt) + P(losing in 1st attempt )*P(winning in 2nd attempt) + P(losing in 1st & 2nd attempt) * P(winning in 3rdattempt)... so on till infinity.

Question asks for Probability of winning after playing the game once, which can be calculated as:

Probability Asked = 1 - Prob of winning in first - probability of always losing (shall converge to zero over time)

=> 1 - 5/9 - ~0
=> 4/9

I may be overthinking, but pls clarify.

Thanks in advance

Bunuel · Answer

Yes, you are overthinking it. Please check the above discussion for the correct interpretation.

sujoykrdatta · Answer

Method 1: 
Prob that neither of the two happens is 4/6 = 2/3
This, prob that neither happens in both cases = 2/3 × 2/3 = 4/9
Thus, prob that at least one happens = 1 - 4/9 = 5/9

Method 2: 
Prob that at least one happens 
= Gets on the first but not other + gets on the other but not on the first + gets on both
= 2/6 × 4/6 + 4/6 × 2/6 + 2/6 × 2/6
= 20/36 
= 5/9

Ans C

billlyjoehbs · Answer

Bunuel

Thanks for your prompt feedback. While the question otherwise is very easy to solve, the language still feels trippy when I read that part. The usage feels unorthodox. Usually, the GMAT question stems ask for probability of winning  in  the first attempt rather than say  after  first attempt. Just want to get this right to avoid any such future mistakes.

Thanks again!

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