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Question: Anand plays with Karpov \(3\) games of chess. The probability that he wins a game is \(0.5\), loses ...

Anand plays with Karpov 33 games of chess. The probability that he wins a game is 0.50.5, loses with probability 0.30.3 and ties with probability 0.20.2. If he plays 33 games then find the probability that he wins at least two games:
A.12\dfrac{1}{2}
B.13\dfrac{1}{3}
C.14\dfrac{1}{4}
D.15\dfrac{1}{5}

Explanation

Solution

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 33 games of chess.
Let the probability of winning the game is pp.
According to our question the probability that he wins a game is 0.50.5, therefore, p=0.5p = 0.5
Let qqdenote the probability of not winning a game it means, either tie or lose the game.
According to our question loses with probability 0.30.3 and ties with probability 0.20.2, therefore, q=0.2+0.3=0.5q = 0.2 + 0.3 = 0.5
Or we can say that 1−p1 - palso denotes the probability of not winning.
Therefore, q=1−p=0.5q = 1 - p = 0.5
Now, we use Bernoulli’s trial formula which states that If A denotes the number of successes in nn Bernoulli’s trial with success probability pp and the probability of getting failure is 1−p1 - p then probability distribution of the random variable XX is defined as,
P(X=x)=nCxpx(1−p)n−xP(X = x) = {}^n{C_x}{p^x}{(1 - p)^{n - x}}, x=0,1,2,....nx = 0,1,2,....n
In our case the value of xxis 22 and 33 as he has to win at least two games.
Therefore, probability is defined as P=3C2p2(1−p)+3C3p3(1−p)0P = {}^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 nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
Therefore,
P=3!2!1!(0.5)20.5+3!3!0!(0.5)3 P=3(0.5)3+(0.5)3 P=4(0.5)3 P=12  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 12\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.