Conditional Probability

Problem 108: Heads and Tails Three ordinary quarters and a fake quarter with two heads are placed in a hat. One quarter is selected at random and tossed twice. If the outcome is “HH”, what is the probability that the fake quarter was selected?

Solution: I believe that sometimes is better to start from the end when solving a mathematics problem. Here we are asked to compute the probability that the selected coin is fake given that we get “HH” as the outcome. That is,

(1) $\begin{equation*} \text{P(FAKE$|$ HH)} = \frac{\text{P(FAKE and HH)}}{\text{P(HH)}}.\end{equation*}$

Let’s calculate the probabilities separately.

i) Since there are 4 quarters, 1 fake and 3 non fake, the probability of choosing the fake coin is $1/4$ . If we toss the fake coin twice, the probability of obtaining “HH” is 1 or 100% since it contains 2 heads and no tail side. This means that

(2) $\begin{equation*} \text{P(FAKE and HH)} = \frac{1}{4} \cdot 1 = \frac{1}{4}.\end{equation*}$

ii) Similarly, since there are 4 quarters, 1 fake and 3 non fake, the probability of choosing the fake coin is $3/4$ . If we toss the fake coin twice, the probability of obtaining “HH” is 1 or 100% since it contains 2 heads and no tail side. Morever, if we toss the non fake coin twice, the probability of obtaining “HH” is $1/4$ . That is,

(3) $\begin{equation*} \text{P(HH)} = \frac{1}{4} \cdot 1 + \frac{3}{4} \cdot \frac{1}{4}.\end{equation*}$

Therefore,

(4) $\begin{equation*} \text{P(FAKE$|$ HH)} = \frac{1/4}{1/4(1+3/4)}= \frac{4}{7} \approx 0.57142857.\end{equation*}$

Hence, there is approximately a 57% chance that the fake quarter was selected.

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