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

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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}$

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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.

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