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

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01 May 2013, 10:03

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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?

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

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01 May 2013, 10:40

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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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01 May 2013, 10:41

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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 \(\frac{2}{9}+\frac{2}{9}+\frac{1}{9}=\frac{5}{9}\)
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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

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

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

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05 May 2013, 04:59

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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05 May 2013, 05:43

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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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05 May 2013, 08:09

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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

Instead of trying to count the overlapping events and thereby complicating the probability calculation, we can simply calculate the probability of 'not winning' and subtract it from 1 to get the probability of 'winning'

Therefore required probability \(P = 1 - (\frac{4}{6})*(\frac{4}{6})\)

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

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30 Jul 2013, 10:51

I also got confused with this at least, i interpreted it as 5-x 6-x 1-x 5-1 6-1 vice-versa cases.
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]

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03 Aug 2013, 03:10

dyaffe55 wrote:

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

I made the same mistake as well... But for the condition above wouldnt the answer be 5/6?

Bunuel/Zarroulou could you confirm? If a win was 1,5,6 instead of 1 and 5?
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]

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03 Aug 2013, 03:52

I meant the condition for a win was at least a five (5 or 6) or 1 on either die...
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Re: In a game, one player throws two fair, six-sided die at the [#permalink]

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23 Mar 2014, 16:14

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Question is not clear when it refers to 'at least' does it refer to at least a 5, that meaning 5 or 6? Or does it refer to have at least one of two: either a 5 or a 1?

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

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20 Jun 2015, 11:24

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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

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

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04 Dec 2015, 17:06

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.

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