Let A and B be two events such that \(\mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{2}{5}\), \(\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1}{7}\) and \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{9}\). Consider (S1) \(\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\frac{5}{6}\), (S2) \(\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\frac{1}{18}\). Then

(A) Both (S1) and (S2) are true

(B) Both (S1) and (S2) are false

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Probability

Let A and B be two events such that

P (B ∣ A) = \frac{2}{5}

P (A ∣ B) = \frac{1}{7}

and

P (A \cap B) = \frac{1}{9}

. Consider
(S1)

P (A^{'} \cup B) = \frac{5}{6}

,
(S2)

P (A^{'} \cap B^{'}) = \frac{1}{18}

. Then

Answer: (A)

From

P (A ∣ B) = \frac{1}{7}

and

P (A \cap B) = \frac{1}{9}

, we get

P (B) = \frac{P ( A \cap B )}{P ( A ∣ B )} = \frac{1/9}{1/7} = \frac{7}{9}

. From

P (B ∣ A) = \frac{2}{5}

and

P (A \cap B) = \frac{1}{9}

, we get

P (A) = \frac{P ( A \cap B )}{P ( B ∣ A )} = \frac{1/9}{2/5} = \frac{5}{18}

. For (S1):

P (A^{'} \cup B) = 1 - P (A \cap B^{'}) = 1 - [P (A) - P (A \cap B)] = 1 - \frac{5}{18} + \frac{1}{9} = \frac{15}{18} = \frac{5}{6}

. For (S2):

P (A^{'} \cap B^{'}) = 1 - P (A \cup B) = 1 - [P (A) + P (B) - P (A \cap B)] = 1 - \frac{5}{18} - \frac{7}{9} + \frac{1}{9} = \frac{1}{18}

. Both statements are true.

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option