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Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.

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Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.  [#permalink]

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New post 26 Jan 2019, 22:59
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Question Stats:

39% (02:33) correct 61% (02:27) wrong based on 38 sessions

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Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze. They all toss each of their coins once and record the results. Two results are the same if two gold coins land on the same face, two silver coins land on the same face, and two bronze coins land on the same face. What is the probability that Paul's result is the same as at least one of the other two results?

A. 15/32
B. 15/64
C. 31/64
D. 3/8
E. 1/2
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Re: Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.  [#permalink]

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New post 27 Jan 2019, 00:13
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dreamerchaser, what is the question source?

It's very easy to get confused in this question.

Each coin can be heads or tails. Each person has 3 coins. So there are 8 combinations of results for each person \((2^3)\). We are not trying to find find when any of their combinations match, only when P matches Q or R.

I am answering for the 1-p, so the probability of P not matching Q or R.
P is guaranteed to get the result of P, so we can ignore it.

Q can get 7 combinations different to P's (7/8) and R can get 7 different combinations too (7/8) - remember, we don't have to find the combinations that P and Q don't have.

\(\frac{7}{8}·\frac{7}{8}=\frac{49}{64}\)
Is the probability that P won't match Q or R.

1-p is the probability that it will
\(\frac{64}{64}-\frac{49}{64}=\frac{15}{64}\)

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Re: Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.  [#permalink]

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New post 27 Jan 2019, 04:32
still not sure if I understood the explanation.
Can any expert comment on it
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Re: Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.  [#permalink]

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New post 27 Jan 2019, 09:45
Anyone can help to solve this question in an understandable way :)
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Re: Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.  [#permalink]

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New post 01 Mar 2019, 07:00
dreamerchaser wrote:
Anyone can help to solve this question in an understandable way :)

Let me try:

To answer the question we need to find:
Number of occurrences when Paul's result is same as Richard's alone, or Quinn's alone . "Or" is used because question stem is "What is the probability that Paul's result is the same as at least one of the other two results?". Had the question been "What is the probability that Paul's result is the same as two other players'?" we needed to calculate different number.
The probability that Event A (Paul's toss is the same as Richard's) or Event B (Paul's toss is the same as Quinn's) occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur (that is Paul's toss is the same as Q's and R's ).


Lets start
1. When tossed, a coin lands either heads(H) or tails(T), so 2 options exist per a single coin toss.
2. When 3 coins are tossed simultaneously, number of possible different combinations = 2*2*2 = 2^3= 8. All possible different combinations (letters correspond to Gold, Silver and Bronze coins respectively):
HHH
HHT
HTH
HTT
THH
TTH
TTT
THT

3. We need not to calculate how many combinations Paul will have when he tosses his coins, because for any combination of his coins, number of different combinations for R and Q is always the same and equal to 8.

Therefore,
1.probability that Paul's toss is the same as R's = 1/8
2. probability that Paul's toss is the same as Q's = 1/8
3. probability that Paul's toss is the same as Q's and R's = ? Now to answer this question we need to find how many different combinations R's and Q's could have = 8*8 = 64, therefore, the probability = 1/64

The probability that Event A or Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur.
=> 1/8+1/8-1/64 = 15/64

Different approach would be to calculate probability of that Paul's toss is not the same as either R's or Quinn =event C and subtract C from 1 to get the answer. (1-eventC), which is shown by philipssonicare
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Re: Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.   [#permalink] 01 Mar 2019, 07:00
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Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.

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