Probability — Class 12 Maths Solution

ncert exercise SA NCERT,EX.13.4,Q.10,Page.571
Question

Find the mean number of heads in three tosses of a fair coin.

Step-by-step Solution

.: Sample space $= \{ HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$

Let $X$ denotes the random variable which represents the number of heads in three tosses of a fair coin.

$\therefore \quad X$ can assume values 0, 1, 2 and 3

$\therefore \quad P(X = 0) = \cfrac{1}{8};P(X = 1) = \cfrac{3}{8};P(X = 2) = \cfrac{3}{8}$ and $P(X = 3) = \cfrac{1}{8}$

Hence, the probability distribution :

figure

$\therefore \quad$

Mean $= E(X) = \Sigma xp(x) = 0 \times \cfrac{1}{8} + 1 \times \cfrac{3}{8} + 2 \times \cfrac{3}{8} + 3 \times \cfrac{1}{8}$

$= 0 + \cfrac{3}{8} + \cfrac{6}{8} + \cfrac{3}{8} = \cfrac{{12}}{8} = \cfrac{3}{2}$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Probability. Curated by Sachin Sharma. Free for all students.