Probability — Class 12 Maths Solution

Here, $S =${(1,2),(2,1),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1),(1,5),(5,1),(2,4),(4,2),

(2,5),(5,2),(3,4),(4,3),(3,5),(5,3),(5,4),(4,5)}

$\Rightarrow$ $n(S) = 20$

Let random variable be $X$ which denotes the sum of the numbers on two cards drawn.

$\therefore$ $X = 3,4,5,6,7,8,9$
At $X = 3,P(X) = \frac{2}{{20}} = \frac{1}{{10}}$

At $X = 4,P(X) = \frac{2}{{20}} = \frac{1}{{10}}$

At $X = 5,P(X) = \frac{4}{{20}} = \frac{1}{5}$

At $X = 6,P(X) = \frac{4}{{20}} = \frac{1}{5}$

At $X = 7,P(X) = \frac{4}{{20}} = \frac{1}{5}$

At $X = 8,P(X) = \frac{2}{{20}} = \frac{1}{{10}}$

At $X = 9,P(X) = \frac{2}{{20}} = \frac{1}{{10}}$

$\therefore$ $\quad$

Mean, $E(X) = \Sigma XP(X) = \frac{3}{{10}} + \frac{4}{{10}} + \frac{5}{5} + \frac{6}{5} + \frac{7}{5} + \frac{8}{{10}} + \frac{9}{{10}}$

$= \frac{{3 + 4 + 10 + 12 + 14 + 8 + 9}}{{10}} = 6$

Also, $\Sigma {X^2}P(X) = \frac{9}{{10}} + \frac{{16}}{{10}} + \frac{{25}}{5} + \frac{{36}}{5} + \frac{{49}}{5} + \frac{{64}}{{10}} + \frac{{81}}{{10}}$

$= \frac{{9 + 16 + 50 + 72 + 98 + 64 + 81}}{{10}} = 39$

$\therefore$ ${\mathop{\rm Var}\nolimits} (X) = \Sigma {X^2}P(X) - {\left[ {\sum X P(X)} \right]^2}$

$= 39 - {(6)^2} = 39 - 36 = 3$