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Problem: Five pennies are sitting on a table. One is a trick coin that has heads on both sides, but the other four are normal. You pick up a penny at random and flip it four times, getting heads each time. Given this, what is the probability that you picked up the two-headed penny?

Attempt at Problem:

1) What is the probability that I pick up the double heads penny --> (1/5)

2) What is the probability that I flipped four heads if I picked up a normal coin? --> (1/2)^4 = (1/16)

Question: I am unsure of how to solve. Could someone show me how you incorporate these two parts to find if you picked up the two-headed penny?

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First, a rough idea.

See, you did not see the coin which you flipped, but have got all four heads. What you definitely know, is that the coin you chose, is either one of the four "fair" pennies, or the "trick" penny.

So, all you have to do is this :

  • The probability of choosing the trick penny is $\frac 15 = 0.2$. If you did choose the trick penny, then the probability of getting four heads is $\mathbf{1}$, since both sides of the penny are heads.

  • The probability of picking a "fair" penny is $\frac 45 = 0.8$, and if you did choose one of these fair coins, then the probability of getting four heads is $\frac 1{16} = 0.0625$.

But either one of the two possibilities has definitely occurred, since we know we have seen four heads. So, the answer is "the probability that you chose the trick coin and got four heads" divided by "the probability that you just get four heads, regardless of which coin you chose", which is : $\frac{0.2 \times 1}{0.2 \times 1 + 0.8 \times 0.0625} = \frac{0.2}{0.25} = \frac 45$.


A more mathematical interpretation would go like this : let $A$ be the event that "four heads appear", let "B" be the event that the "trick penny" was chosen. Then, we want to calculate $P(B | A)$. We write: $$ P(B | A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A \cap B)}{P(A \cap B) + P(A \cap B^c)} $$

Now, $P(A \cap B)$ is the probability of choosing the trick coin and getting four heads, while $P(A \cap B^c)$ is the probability of choosing any fair coin and getting four heads. Now the logic used in the "rough idea" follows, giving the answer $\frac 45$.

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