A box contains three coins: a fair coin, a coin with a head on both sides, and a biased coin that shows a head with probability \(\displaystyle\frac{1}{4}\). A coin is chosen at random from the box and tossed twice. If both tosses show a head, then the probability that the chosen coin is the one with a head on both sides is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{16}{21}\)
Let \(E_1, E_2, E_3\) be the events that the fair coin, the two-headed coin and the biased coin are chosen, each with probability \(\frac{1}{3}\), and let A be the event of two heads. The two tosses of the chosen coin are independent, so
\( \displaystyle P(A|E_1) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
\(P(A|E_2) = 1\)
\( \displaystyle P(A|E_3) = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)
By Bayes’ theorem (the factors \(\frac{1}{3}\) cancel):
\( \displaystyle P(E_2|A) = \frac{1}{\frac{1}{4}+1+\frac{1}{16}} \)
\( \displaystyle P(E_2|A) = \frac{1}{\frac{21}{16}} = \frac{16}{21} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{4}{7}\) — Incorrect. This is the answer for a single head, \(1\div\left(\frac{1}{2}+1+\frac{1}{4}\right)\); the second head makes the two-headed coin even more likely.
Option (3) \(\displaystyle\frac{1}{3}\) — Incorrect. This is the chance of choosing that coin before any toss.
Option (4) \(\displaystyle\frac{4}{21}\) — Incorrect. This is \(\frac{1}{4}\div\frac{21}{16}\), the chance that the fair coin was chosen.