Question

Anand plays with Karpov $3$ games of chess. The probability that he wins a game is $0.5$, loses with probability $0.3$ and ties with probability $0.2$. If he plays $3$ games then find the probability that he wins at least two games:
A.$\dfrac{1}{2}$
B.$\dfrac{1}{3}$
C.$\dfrac{1}{4}$
D.$\dfrac{1}{5}$

Answer 1

Hint: At least two wins means either two wins or three wins. Write the winning, loose and tie probability and use Bernoulli's trial to evaluate the probability of winning at least two games.

Complete step-by-step answer:
We are given that Anand plays with Karpov $3$ games of chess.
Let the probability of winning the game is $p$.
According to our question the probability that he wins a game is $0.5$, therefore, $p = 0.5$
Let $q$denote the probability of not winning a game it means, either tie or lose the game.
According to our question loses with probability $0.3$ and ties with probability $0.2$, therefore, $q = 0.2 + 0.3 = 0.5$
Or we can say that $1 - p$also denotes the probability of not winning.
Therefore, $q = 1 - p = 0.5$
Now, we use Bernoulli’s trial formula which states that If A denotes the number of successes in $n$ Bernoulli’s trial with success probability $p$ and the probability of getting failure is $1 - p$ then probability distribution of the random variable $X$ is defined as,
\[P(X = x) = {}^n{C_x}{p^x}{(1 - p)^{n - x}}\], $x = 0,1,2,....n$
In our case the value of $x$is $2$ and $3$ as he has to win at least two games.
Therefore, probability is defined as $P = {}^3{C_2}{p^2}(1 - p) + {}^3{C_3}{p^3}{(1 - p)^0}$
Substitute all the values on the given formula and evaluate the probability.
Use the formula ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Therefore,
$
  P = \dfrac{{3!}}{{2!1!}}{(0.5)^2}0.5 + \dfrac{{3!}}{{3!0!}}{(0.5)^3} \\
  P = 3{(0.5)^3} + {(0.5)^3} \\
  P = 4{(0.5)^3} \\
  P = \dfrac{1}{2} \\
 $
Hence, the probability that he wins at least two games is $\dfrac{1}{2}$
Therefore, option (A) is correct.

Note: We understand more about Bernoulli's trial.
It is one of the simplest experiments you can do in probability because in this experiment we have one of the two possible outcomes.
For example, when we toss a coin, there is only one possibility out of the two that is either it will be head or tail.