Question

A player randomly chooses one of the coin $A,B$. Coin $A$ has a probability of heads $\frac{3}{4}$ and coin $B$ has a probability of heads $\frac{1}{4}$. He tosses the coin twice. The probability that he obtains heads only once is

• A. $\frac{3}{8}$
• B. $\frac{5}{8}$
• C. $\frac{7}{8}$
• D. None of these

Answer 1

The correct option is A

$\frac{3}{8}$

Explanation for the correct option:

Step 1: Find the probability of the head comes on Coin $A$:

Find the probability that coin $A$ is chosen and the head comes in it.

$\begin{array}{rcl}P\left(H\right)& =& P\left(A\right)·P\left(\frac{H}{A}\right)\\ & =& \frac{1}{2}·\left(2·\frac{3}{4}·\frac{1}{4}\right)\\ & =& \frac{3}{16}\end{array}$

Step 2: Find the probability of the head comes on Coin $B$:

Find the probability that coin $B$ is chosen and the head comes in it.

$\begin{array}{rcl}P\left(H\right)& =& P\left(B\right)·P\left(\frac{H}{B}\right)\\ & =& \frac{1}{2}·\left(2·\frac{1}{4}·\frac{3}{4}\right)\\ & =& \frac{3}{16}\end{array}$

Step 3: Find the probability that head comes on tossing any coin:

Sum up the probability of getting head on any of the coins to get the required probability.

$\begin{array}{rcl}P\left(H\right)& =& \frac{3}{16}+\frac{3}{16}\\ & =& \frac{3}{8}\end{array}$

Hence, option (A) is the correct answer.