Question

There are 3 coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

Short answer

With the help of diagram below we get the answer as follows

$\frac{1}{3}\times 1=\frac{1}{3}$

$\frac{1}{3}\times \frac{1}{2}=\frac{1}{6}$

$\frac{1}{3}\times \frac{3}{4}=\frac{1}{4}$

Step-by-step solution

Given information

From the given information that there are three coins in a box. One is two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time.

One of the 3 coins is selected at random and flipped, it shows heads.

We have to find the probability that it was the two-headed coin.

Diagram

Let us assume that $T=$two-headed;

$F=$fair, $B=$biased; head.

We need to construct the following diagram is as follows:

Final answer

From the diagram we have

$\frac{1}{3}\times 1=\frac{1}{3}$

$\frac{1}{3}\times \frac{1}{2}=\frac{1}{6}$

$\frac{1}{3}\times \frac{3}{4}=\frac{1}{4}$