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

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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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karishmatandon wrote:
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

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

Cheers,
Brent
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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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
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
Zarrolou wrote:
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 $$\frac{2}{9}+\frac{2}{9}+\frac{1}{9}=\frac{5}{9}$$

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?
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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skamal7 wrote:
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
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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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
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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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

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In a game, one player throws two fair, six-sided die at the [#permalink]
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.

Originally posted by whitehalo on 04 Dec 2015, 18:06.
Last edited by whitehalo on 22 Jun 2016, 15:13, edited 1 time in total.
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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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).

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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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kalcee wrote:
I too got confused with the word "at least" I assumed that either a 1, 5 or 6 would constitute a win. Hmmm

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.
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
karishmatandon wrote:
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

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
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
Archit3110 wrote:
karishmatandon wrote:
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

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
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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AnirudhaS wrote:
Archit3110 wrote:
karishmatandon wrote:
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

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

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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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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 = $$\frac{20}{36}$$ = $$\frac{5}{9}$$

Hope it helps!

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

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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
I agree with ShukhratJon. This particular question is not properly worded.
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]
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