Question

If $ A $ and $ B $ are independent events associated to some experiment $ E $ such that $ P\left(A^{c}\cap B\right) = \frac{2}{15} \,$ and $\, P\left(A\cap B^{c}\right)=\frac{1}{6}$, then $P\left(B\right) $ , then $ P(B) $ is equal to

• A. $ \frac{1}{6},\frac {1}{5} $
• B. $ \frac{1}{6},\frac {4}{5} $
• C. $ \frac{4}{5},\frac {1}{5} $
• D. $ \frac{4}{5},\frac {5}{6} $

Answer 1

Answer: (B)

We have, $P(A^c \cap B) = \frac{2}{15} $ and $P(A\cap B^c) = \frac{1}{6}$
$\Rightarrow P(A^c) \cdot P(B) = \frac{2}{15} $ and $P(A) \cdot P(B^c) = \frac{1}{6}$
$\Rightarrow [ 1 - P(A) \cdot P(B) = \frac{2}{15}$ and $P(A) \cdot [1 - P(B)] = \frac{1}{6}$
$\Rightarrow [ 1 - P(A)] \cdot P(B) = \frac{2}{15} $ and $P(A) = \frac{1}{6[1-P(B)]}$
$\therefore [ 1 - \frac{1}{6[1-P(B)]}] \cdot P(B) = \frac{2}{15}$
$\Rightarrow \frac{[6 - 6P(B) -1]}{6[1 - P(B)]} \cdot P(B) = \frac{2}{15}$
$\Rightarrow 15 [ 5 - 6\,P(B)] P(B) = 2 \times 6 [ 1 - P(B) ]$
$ \Rightarrow 75\,P(B) - 90\,P(B)^2 = 12 - 12\,P(B)$
Let $P(B) = x$
$\therefore 75x - 90 x^2 = 12 - 12 x$
$\Rightarrow 90x^2 - 75 x - 12 x + 12 = 0$
$\Rightarrow 90x^2 - 87 x + 12 = 0$
$\Rightarrow 90\,x^2 - 15 x -72 x + 12 = 0$
$\Rightarrow 15 x (6x -1) - 12(6x - 1) = 0$
$\Rightarrow (6x -1)(15x - 12) = 0$
$\Rightarrow 6x -1 = 0 $ or $15x -12 = 0$
$\Rightarrow x = \frac{1}{6}$ or $x = \frac{12}{15} = 0$
$\Rightarrow x = \frac{1}{6}$ or $ x = \frac{4}{5}$