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

Question

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.

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

Source: Official Guide

ceilidh.erickson · Answer

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
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

BTG Expert Team · Answer

What we see: a Two-Part Analysis question about coins. We pick one value for $M$ and one for $N$ so that $\frac{1}{p} + \frac{1}{q} = 72$ and $M < N$. Step 1. Write $p$ and $q$ in terms of $M$ and $N$. What we see: each fair coin lands faceup with chance $\frac{1}{2}$, and the coins do not affect each other. What it means: the chance that all $M$ coins land faceup is $\frac{1}{2}$ multiplied by itself $M$ times, so $p = \frac{1}{2^M}$. In the same way, $q = \frac{1}{2^N}$. What we do and why: we flip each fraction, because the equation uses $\frac{1}{p}$ and $\frac{1}{q}$: $\frac{1}{p} = 2^M$ and $\frac{1}{q} = 2^N$. Step 2. Turn the condition into a sum of powers of 2. What we see: $\frac{1}{p} + \frac{1}{q} = 72$. What it means: $2^M + 2^N = 72$. We need two powers of 2 that add up to 72. What we do and why: list the powers of 2 below 72: 1, 2, 4, 8, 16, 32, 64. Try the biggest first. $64 + 8 = 72$ works. Check the rest: $72 - 32 = 40$, $72 - 16 = 56$, $72 - 4 = 68$, $72 - 2 = 70$, $72 - 1 = 71$. None of those is a power of 2. So only one pair works: 8 and 64. Step 3. Match the pair to $M$ and $N$. What we see: $8 = 2^3$ and $64 = 2^6$, and the question says $M < N$. What it means: $M = 3$ and $N = 6$. What we do and why: check: $p = \frac{1}{8}$, $q = \frac{1}{64}$, and $8 + 64 = 72$. Since only one pair of powers works, this is the only answer, whatever other values the table lists. The trap: treating the chance for $M$ coins as $\frac{M}{2}$, or adding $p + q$ instead of $\frac{1}{p} + \frac{1}{q}$. Both lead to numbers that never reach 72. Another slip is swapping the columns: $M = 6$, $N = 3$ breaks $M < N$. Faster way: write 72 as a sum of powers of 2 by taking the biggest one that fits: $72 = 64 + 8$. That split is always the only one. Takeaway: the chance that $n$ fair coins all land the same chosen way is $\frac{1}{2^n}$, so its flip is $2^n$. When a question adds flipped probabilities, look for powers of 2. Answer: M 3, N 6

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