Two independent events A and B are such that P(A B)=(2)/(3) and P(A B)=(1)/(6) . If P(B) < P(A) , then what is P(B) equal to ?

Correct answer: B. (1)/(3) P(A B)=(2)/(3)P(A B)=(1)/(6) Since, A and B are independent events, P(A B)=P(A) · P(B)=(1)/(6) ...(1) aligned & P(A B)=(2)/(3) P(A)+P(B)-P(A B)=(2)/(3) & P(A)+P(B)-(1)/(6)=(2)/(3) aligned P(A)+P(B)=(2)/(3)+(1)/(6)=(5)/(6) ...(2) aligned & from (1), (2), P(B)=(1)/(3) or (1)/(2) & P(B) < P(A), P(B)=(1)/(3) aligned

NDA 2019

Exam: NDA
Year: 2019
Subject: Mathematics
Chapter: Probability

NDA 2019 · Mathematics · Probability

Two independent events A and B are such that P(A B)=(2)/(3) and P(A...

Two independent events

A

and

B

are such that

P(A \cup B)=\frac{2}{3}

and

\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}

. If

\mathrm{P}(\mathrm{B}) < \mathrm{P}(\mathrm{A})

, then what is

\mathrm{P}(\mathrm{B})

equal to ?

Options

A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{1}{6}$

Correct answer

B. $\frac{1}{3}$

Step-by-step solution

P(A \cup B)=\frac{2}{3}P(A \cap B)=\frac{1}{6}

Since, A and B are independent events,

P(A \cap B)=P(A) \cdot P(B)=\frac{1}{6}

...(1)

\begin{aligned} & P(A \cup B)=\frac{2}{3} \Rightarrow P(A)+P(B)-P(A \cap B)=\frac{2}{3} \\ & \Rightarrow P(A)+P(B)-\frac{1}{6}=\frac{2}{3}\end{aligned}

\Rightarrow P(A)+P(B)=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}

...(2)

\begin{aligned} & \text{ from (1), (2), } \mathrm{P}(\mathrm{B})=\frac{1}{3} \text{ or } \frac{1}{2} \\ & \because P(B) < P(A), P(B)=\frac{1}{3}\end{aligned}

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