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A fair coin has 2 distinct flat sides—one of which bears the image of

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A fair coin has 2 distinct flat sides—one of which bears the image of  [#permalink]

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New post 25 Aug 2018, 04:32
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A fair coin has 2 distinct flat sides—one of which bears the image of a face and the other of which does not—and when the coin is tossed, the probability that the coin will land faceup is ½. For certain values of M, N, p, and q, when M fair coins are tossed simultaneously, the probability is p that all M coins land faceup, and when N fair coins are tossed simultaneously, the probability is q that all N coins land faceup. Furthermore, M < N and 1/p + 1/q = 72.

In the table, select a value for M and a value for N that are jointly consistent with the given information. Make only two selections, one in each column.



OA:
M=3
N=6
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Re: A fair coin has 2 distinct flat sides—one of which bears the image of  [#permalink]

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New post 25 Aug 2018, 18:47
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Blackishmamba wrote:
A fair coin has 2 distinct flat sides—one of which bears the image of a face and the other of which does not—and when the coin is tossed, the probability that the coin will land faceup is ½. For certain values of M, N, p, and q, when M fair coins are tossed simultaneously, the probability is p that all M coins land faceup, and when N fair coins are tossed simultaneously, the probability is q that all N coins land faceup. Furthermore, M < N and 1/p + 1/q = 72.

In the table, select a value for M and a value for N that are jointly consistent with the given information. Make only two selections, one in each column.



OA:
M=3
N=6


Probability that all M have same outcome = 1/2^m
And probability that all N have same outcome = 1/2^n
So 1/(1/2)^m + 1/(1/2)^n = 72 thus 2^m+2^n=72
2^m(1+2^{n-m})=72=2^3*3^2... So m=3
So 1+2^{n-m} must be a multiple of 9.. so 2^{n-m}=8....
n-m=3...n=6
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html


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Re: A fair coin has 2 distinct flat sides—one of which bears the image of  [#permalink]

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New post 10 Sep 2018, 19:19
1
Trial and error would be quicker.

Probability for each event irrespective of the coin ) M or N is 1/2

i.e. if M =1 coin
> p (probability of face) =1/2

Similarly for N

Now for M coins
> p= (1/2)^M

For M coins
> p= (1/2)^N

Now to confuse us , the reciprocal of sums is given so that we get confused.
But one clue is that the sum is 72, an integer, so you need not fear.

1/p +1/q =72

1/(1/2)^M + 1/(1/2)^N = 72

>> 2^M +2^N = 72.

All you have to do list down some powers of two.not much just few.

2
4
8
16
32
64

Given M<N

Trial and error gives us
8+64 =72
2^3 + 2^6 = 72

M=3 N=6

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Re: A fair coin has 2 distinct flat sides—one of which bears the image of  [#permalink]

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New post 17 Sep 2018, 03:55
1
Another way would be trial basis...

By solving the given equation, we can say p+q=72pq or q=p/(72p-1)

We also know p and q will have denominators in the form of 2^x

Start with option A)
If M=2, then
p= (1/2)^2 = 1/4 -> q=1/88 - not possible as denominator not of the form 2^x

Similarly, Option B)
If M=3, then
p=1/8 -> q=1/64 = 1/(2^6)
Therefore N = 6.

Answer:
M = 3 (Option B)
N = 6 (Option E)

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Re: A fair coin has 2 distinct flat sides—one of which bears the image of   [#permalink] 17 Sep 2018, 03:55
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