A fair coin has 2 distinct flat sides-one of which bears t

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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 \(\frac{1}{2}\). 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 face up. Furthermore, \(M < N\) and \(\frac{1}{p} + \frac{1}{q} = 72\).

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.

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The OA is [spoiler]M=3 N=6[/spoiler]

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by ceilidh.erickson » Sat Jun 01, 2019 9:58 am
If one coin flip has a 1/2 chance of landing face up, then the probability that two coin flips will both be face up is \(\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\)

Thus, for any M coin flips, the probability of all flips landing face up is \(p=\frac{1}{2^M}\) , and for any N coin flips, the probability of all flips landing face up is \(q=\frac{1}{2^N}\) .

We're told that \(\frac{1}{p} + \frac{1}{q} = 72\) , so we can infer that \({2^M}+{2^N}=72\) . Think of powers of 2 that would add to 72, starting with the largest power of 2 less than 72:
\({2^6}\) --> 64 + 8 = 72 --> yes! These are both powers of 2.
\({2^5}\) --> 32 + 40 = 72 --> no, 40 is not a power of 2.
\({2^4}\) --> 16 + 56 = 72 --> no, 56 is not a power of 2.
\({2^3}\) --> 8 + 64 = 72 --> yes, we already know this works.
\({2^2}\) --> 4 + 68 = 72 --> no, 68 is not a power of 2.
\({2^1}\) --> 2 + 70 = 72 --> no, 70 is not a power of 2.

The only two numbers that work are \({2^6}\) and \({2^3}\) , 64 and 8. Since we're given the constraint that M must be less than N, \({2^M}\) must equal \({2^3}\) and \({2^N}\) = \({2^6}\) . Thus, [spoiler]M=3 and N=6[/spoiler] .
Ceilidh Erickson
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