A person draws two cards successively with out replacement from a pack of 52 cards. He tells that both cards are aces. The probability that both are aces if there are 60% chances that he speaks truth is equal to:
Answer & explanation
Correct answer: option 1
A1 : both are aces
A2 : both are non-aces
A3 : One ace and other non - ace
E : Person tells that both are aces
$P(E) = P(A_1 ∩ E) + P(A_2 ∩ E) + P(A_3 ∩ E)$
$= P(A_1)P(E/A_1) + P(A_2)P(E/A_2) + P(A_3)P(E/A_3)$
= P (both aces) P(Person speaks truth) + P (both not aces) P (Person speaks lie) + P(One ace and One non-ace) P (Person speaks lie)
$=(\frac{4}{52}×\frac{3}{51})×\frac{6}{10}+\frac{48}{52}×\frac{47}{51}×\frac{4}{10}+2×\frac{48}{52}×\frac{4}{51}×\frac{4}{10}$
$=\frac{4×3×6+48×47×4+2×48×4×4}{52×51×10}$
$P(A_1/E)=\frac{P(A_1)P(E/A_1)}{P(E)}=\frac{4×3×6}{4×3×6+48×47×4+2×48×4×4}=\frac{3}{443}$
