A multiple-choice question has 5 options of which only one is correct. If a student does home work, then he is sure to identify the correct answer, otherwise the answer is chosen at random. Let A be the event that the student does his home work and B be the event that the student answers correctly. If P(A)=(2/3) , then P((A/B)) is equal to

Question

A multiple-choice question has 5 options of which only one is correct. If a student does home work, then he is sure to identify the correct answer, otherwise the answer is chosen at random. Let A be the event that the student does his home work and B be the event that the student answers correctly. If P(A)=(2/3) , then P((A/B)) is equal to (A) (10/11) (B) (4/5) (C) (3/7) (D) (6/7)

Tardigrade Admin · Accepted Answer

P((B/Ac))=(1/5), P(A)=(2/3) P(A / B)=(P(A ∩ B)/P(B))=(P(A) ⋅ P(B / A)/P(A) P((B)A)+P(Ac) P(B / Ac)) =((2/3) × 1/(2)3 × 1+(1)3 × (1/5)=(10/11)

$\frac{10}{11}$

$\frac{4}{5}$

$\frac{3}{7}$

$\frac{6}{7}$

$P (\frac{B}{A ^{c}}) = \frac{1}{5}, P (A) = \frac{2}{3}$
$P (A / B) = \frac{P ( A \cap B )}{P ( B )} = \frac{P ( A ) \cdot P ( B / A )}{P ( A ) P ( \frac{B}{A} ) + P ( A ^{c} ) P ( B / A ^{c} )}$
$= \frac{\frac{2}{3} \times 1}{\frac{2}{3} \times 1 + \frac{1}{3} \times \frac{1}{5}} = \frac{10}{11}$

1110​

54​

73​

76​

P(AcB​)=51​,P(A)=32​ P(A/B)=P(B)P(A∩B)​=P(A)P(AB​)+P(Ac)P(B/Ac)P(A)⋅P(B/A)​ =32​×1+31​×51​32​×1​=1110​

$\frac{10}{11}$

$\frac{4}{5}$

$\frac{3}{7}$

$\frac{6}{7}$

$P (\frac{B}{A ^{c}}) = \frac{1}{5}, P (A) = \frac{2}{3}$
$P (A / B) = \frac{P ( A \cap B )}{P ( B )} = \frac{P ( A ) \cdot P ( B / A )}{P ( A ) P ( \frac{B}{A} ) + P ( A ^{c} ) P ( B / A ^{c} )}$
$= \frac{\frac{2}{3} \times 1}{\frac{2}{3} \times 1 + \frac{1}{3} \times \frac{1}{5}} = \frac{10}{11}$