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 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 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.

Image

OA:
[spoiler]M=3
N=6[/spoiler]

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Ceilidh Erickson
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we have the equation:

1/p + 1/q = 72

To find consistent values for M and N, let's consider some options:

Option 1: M = 1 and N = 2
If we toss 1 fair coin, the probability of it landing faceup is 1/2. Similarly, if we toss 2 fair coins, the probability that both land faceup is (1/2) * (1/2) = 1/4. However, when we substitute these values into the equation, we get 1/1 + 1/1/4 = 1 + 4 = 5, which does not satisfy the equation.

Option 2: M = 2 and N = 4
If we toss 2 fair coins, the probability that both land faceup is (1/2) * (1/2) = 1/4. Similarly, if we toss 4 fair coins, the probability that all 4 land faceup is (1/2) * (1/2) * (1/2) * (1/2) = 1/16. Substituting these values into the equation, we have 1/1/4 + 1/1/16 = 4 + 16 = 20. This satisfies the equation 1/p + 1/q = 72.

Therefore, a consistent selection would be M = 2 and N = 4.