Question

Anand plays with Karpov $3$ games of chess. The probability that he wins a game is $0.5$, looses with probability $0.3$ and ties with probability $0.2$. If he plays $3$ games then find the probability that he wins atleast two games

• A. $1/2$
• B. $1/3$
• C. $1/4$
• D. $1/5$

Answer 1

The correct option is A $1/2$

Let $p$ denote the probability of winning a game.
Hence $p=0.5$
Now let $q$ denote the probability of not wining a game, ie either losing or a tie.
Hence $q=0.3+0.2=0.5$
$=1-q$.
The number of games (number of Bernoulli's trial) are 3.
Therefore the probability of winning atleast 2 games out of 3 is
$={}^3{C}_2{p}^2.q+{}^3{C}_3{p}^3{q}^0$
$=3.(0.5{)}^2.(0.5)+1(0.5{)}^3$
$=\left(0.5{)}^3\right[3+1]$
$=\frac{1}{8}.4$
$=\frac{1}{2}$.