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Probability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability5 Marks
Question
If A and B are two independent events such that $\text{P}(\overline{\text{A}}\cap\text{B})=\frac{2}{15}$ and $\text{P}(\text{A}\cap\overline{\text{B}})=\frac{1}{6}$, then find P(B).
✓

Answer

We are given
$\text{P}(\overline{\text{A}}\cap\text{B})=\frac{2}{15}$
$\text{P}(\text{A}\cap\overline{\text{B}})=\frac{1}{6}$
Since A, B are independent,
$\therefore\ \text{P}(\overline{\text{A}})\text{ P(B)}=\frac{2}{15}\Rightarrow\ [1-\text{P(A)}]\text{P(B)}=\frac{2}{15}\ .....(\text{i})$
and $\text{P(A) }\text{P }(\overline{\text{B}})=\frac{1}{6}\Rightarrow \text{P(A)}[1-\text{P(B)}]=\frac{1}{6}\ .....\text{(ii)}$
From (i) we get
$\text{P(B)}=\frac{2}{15}\times\frac{1}{1-\text{P(A)}}$
Substituting this value in equation (ii) we get,
$\text{P(A)}\Big[1-\frac{1}{15(1-\text{P(A)})}\Big]=\frac{1}{6}$
$\Rightarrow\ \text{P(A)}\Big[\frac{15(1-\text{P(A)})-2}{15(1-\text{P(A)})}\Big]=\frac{1}{6}$
$\Rightarrow 6P(A) (13 - 15P(A)) = 15(1 - P(A))$
$\Rightarrow 2P(A) (13 - 15P(A)) = 5 - 5P(A)$
$\Rightarrow 26P(A) - 30[P (A)]^2 + 5P(A) - 5 = 0$
$\Rightarrow -30[P(A)]^2 + 31P(A) - 5 = 0$
This is a quadrati equation in $x = P(A)$ given as
$-30x^2 + 31x - 5 = 0$
$\Rightarrow 30x^2 - 31x + 5 = 0$
$\therefore\ \text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
Where, $a = +30, b = -31, C = +5$
$\Rightarrow\ \text{x}=\frac{31\pm\sqrt{(-31)^2-4(30)(5)}}{60}$
$=\frac{31\pm\sqrt{961-600}}{60}$
$=\frac{31\pm19}{60}$
$=\frac{50}{60},\frac{12}{60}$
$=\frac{5}{6},\frac{1}{5}$
$\therefore\ \text{P(A)}=\frac{5}{6}\text{ or }\frac{1}{5}$
Now,
$\text{P(A)}[1-\text{P(B)}]=\frac{1}{6}$
Putting $\text{P(A)}=\frac{5}{6}$
$\frac{5}{6}[1-\text{P(B)}]=\frac{1}{6}$
$1-\text{P(B)}=\frac{1}{5}$
$\text{P(B)}=1-\frac{1}{5}$
$\text{P(B)}=\frac{4}{5}$
Putting $\text{P(A)}=\frac{1}{5}$
$\frac{1}{5}[1-\text{P(B)}]=\frac{1}{6}$
$1-\text{P(B)}=\frac{5}{6}$
$\text{P(B)}=1-\frac{5}{6}$
$\text{P(B)}=\frac{1}{6}$
Hence $\text{P(B)}=\frac{4}{5} \text{ or }\frac{1}{6}$

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