Question

Three coins are tossed. Two of them are fair and one is biased so that a head is three times as likely as a tail. Find the probability of getting two heads and a tail

• A. $\frac{7}{16}$
• B. $\frac{3}{4}$
• C. $\frac{9}{16}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{7}{16}$

Let $H$ and $T$ denote head and tails from a unbiased coin and let $H^{\prime }$ and $T^{\prime }$ denote head and tail from the biased coin.
Then for two head in $3$ tosses, it will be
$HH{T}^{\prime },HT{H}^{\prime },TH{H}^{\prime }$
Now $P\left(H^{\prime }\right)=\frac{3}{4}$ and $P\left(T^{\prime }\right)=\frac{1}{4}$
Thus the required probability is
$=\frac{1}{2}\frac{1}{2}\frac{1}{4}+\frac{1}{2}\frac{1}{2}\frac{3}{4}+\frac{1}{2}\frac{1}{2}\frac{3}{4}$
$=\frac{1+3+3}{16}$
$=\frac{7}{16}$.