There are three coins. One is a two-headed coin ( having head on both faces ), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?

$(a)\;\large\frac{20}{47}\qquad(b)\;\large\frac{10}{47}\qquad(c)\;\large\frac{5}{47}\qquad(d)\;\large\frac{12}{47}$

Probability of getting heaf from coin A = 1
Probability of getting heaf from coin $B = \large\frac{3}{4}$
Probability of getting heaf from coin C = 0.6
$\therefore$ probabilty of getting H by coin A is $p \bigg( \large\frac{A}{H} \bigg)$
$p \bigg( \large\frac{A}{H} \bigg) = \large\frac{p \bigg( \large\frac{H}{A} \bigg) . p(A) }{ p \bigg( \large\frac{H}{A} \bigg).p(A)+p \bigg( \large\frac{H}{B} \bigg) .p(A)+p \bigg( \large\frac{H}{C} \bigg) .p(A)}$
Probability of choosing any one coin is
$p(A) = \large\frac{1}{3}, $$p(B) = \large\frac{1}{3} and p(C) = \large\frac{1}{3} p \bigg( \large\frac{H}{A} \bigg)$$ = 1, p \bigg( \large\frac{H}{B} \bigg) $$= \large\frac{3}{4} and p \bigg( \large\frac{H}{C} \bigg)$$ = 0.6$
$\therefore p \bigg( \large\frac{A}{H} \bigg)$$= \large\frac{1. \large\frac{1}{3}}{1. \large\frac{1}{3}+ \large\frac{3}{4} \times \large\frac{1}{3}+ 0.6 \times \large\frac{1}{3}}$
$\large\frac{\large\frac{1}{3}}{\large\frac{1}{3}+\large\frac{1}{4}+0.2}$ $= \large\frac{\large\frac{1}{3}}{\large\frac{1}{3}+\large\frac{1}{4}+\large\frac{1}{5}}$
$= \large\frac{1}{3} \times \large\frac{60}{12}$
$\large\frac{20}{47}$

Let E1 denote the event of getting a two headed coin. let E2 denote the event of getting a biased coin which shows 75% of head. let E3 denote the event of getting a biased coin which shows 40% of tails. let A be the event of getting a two headed coin. P(E1)=1/3 P(E2)=1/3 P(E3)=1/3 P(A|E1)=2/2=1 P(A|E2)=75/100=3/4 P(A|E3)=100-40=60 =60/100=3/5 P(E1|A)=1/3*1/1/3*1+1/3*3/4+1/3*3/5=20/47 THEREFORE,THE OPTION A IS CORRECT.