If A and B are two independent events such that P(not A intersection B) = 2/15 and P(A intersection not - Maths - Probability

Question

If A and B are two independent events such that P(not A
intersection B) = 2/15 and P(A intersection not B) =1/6, then find
P(B)

Lalit Mehra · Accepted Answer

Dear Student!

Given, A and B are two independent events.

∴ $\bar{A}$ and B are independent events and A and $\bar{B}$ are independent events.

$G i v e n, P (\bar{A} \cap B) = \frac{2}{15} ∴ P (\bar{A}) P (B) = \frac{2}{15} ... (1) G i v e n, P (A \cap \bar{B}) = \frac{1}{6} ∴ P (A) P (\bar{B}) = \frac{1}{6} \Rightarrow P (A) [1 - P (B)] = \frac{1}{6} \Rightarrow P (A) = \frac{1}{6 [1 - P (B)]} ... (2) F r o m (1), w e h a v e [1 - P (A)] P (B) = \frac{2}{15} ∴ (1 - \frac{1}{6 [1 - P (B)]}) P (B) = \frac{2}{15} (U \sin g (2)) \Rightarrow \frac{(6 - 6 P (B) - 1)}{6 - 6 P (B)} \times P (B) = \frac{2}{15} \Rightarrow 15 {5 - 6 P (B)} P (B) = 12 {1 - P (B)} \Rightarrow 5 {5 - 6 P (B)} P (B) = 4 (1 - P (B)) \Rightarrow 25 P (B) - 30 (P (B))^{2} = 4 - 4 P (B) \Rightarrow 30 (P (B {))}^{2} - 29 P (B) + 4 = 0 \Rightarrow 30 (P (B))^{2} - 24 P (B) - 5 P (B) + 4 = 0 \Rightarrow 6 P (B) (5 P (B) - 4) - 1 (5 P (B) - 4) = 0 \Rightarrow (5 P (B) - 4) (6 P (B) - 1) = 0 \Rightarrow 5 P (B) - 4 = 0 or 6 P (B) - 1 = 0 \Rightarrow P (B) = \frac{4}{5} or P (B) = \frac{1}{6}$

Thus, P (B) is $\frac{4}{5} or \frac{1}{6}$ .

Cheers!

If A and B are two independent events such that P(not A intersection B) = 2/15 and P(A intersection not B) =1/6, then find P(B)