If A and B are two independent events such that and then find P(A) and P(B).

Engineering

Mathematics

Basic Definitions Related to Probability

Addition Theorem on Probability

Question

If A and B are two independent events such that $P (\bar{A} \cap B) = \frac{2}{15}$ and $P (A \cap \bar{B}) = \frac{1}{6}$ then find P(A) and P(B).

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Given $P (\bar{A} \cap B) = \frac{2}{15} \Rightarrow P (\bar{A}) P (B) = \frac{2}{15}$ --- (1)

Since events A and B are independent, A and B are also independent.
$P (A \cap \bar{B}) = \frac{1}{6} \Rightarrow P (A) P (\bar{B}) = \frac{1}{6}$ --- (2)

As events A and B are independent, A, B are also independent.
Let P(A) = x, P(B) = y
$\Rightarrow P(A) = 1 - x$ and P(B) = 1 – y

From (1) and (2), we get

(1 - x) y = \frac{2}{15}

i.e.,

y - xy = \frac{2}{15}

--- (3)
and

x (1 - y) = \frac{1}{6}

i.e.,

x - xy = \frac{1}{6}

-- (4)
Subtracting (3) and (4), we get

x - y = \frac{1}{6} - \frac{2}{15} \Rightarrow x - y = \frac{1}{30} \Rightarrow x = y + \frac{1}{30}

----(5)

Putting this value of x in (3), we get

y - (y + \frac{1}{30}) y = \frac{2}{15} \Rightarrow \frac{29}{30} y - y^{2} = \frac{2}{15}

\Rightarrow 30y 2 - 29y + 4 = 0 \Rightarrow (5y - 4) (6y - 1) = 0

y = \frac{4}{5}, \frac{1}{6}

From (5), when

y = \frac{4}{5} x = \frac{4}{5} + \frac{1}{30} = \frac{5}{6}

when

y = \frac{1}{6} x = \frac{1}{6} + \frac{1}{30} = \frac{1}{5}

Hence, if

P (A) = \frac{5}{6}

then

and if

P (A) = \frac{1}{5}

, then

P (B) = \frac{1}{6}