Question

A computer chess game and a human chess champion are evenly matched. They play ten games. Find probabilities for the following events. a. They each win five games. b. The computer wins seven games. c. The human chess champion wins at least seven games.

Answer 1

Answer:

(a) The probability that each of them win five games is 0.2461.

(b) The probability that the computer wins seven games is 0.1172.

(c) The probability that the human wins at least 7 games is 0.1711.

Step-by-step explanation:

Let the random variable X = the human wins a chess game.

The probability that the human wins a game is, P (X) = p = 0.50.

The number of games played by the computer and the human is, n = 10.

The random variable $X\sim Bin(n = 10,p=0.50)$

The probability distribution of the Binomial random variable X is:

$P (X=x)={n\choose x}p^{x}(1-p)^{n-x}={10\choose x}(0.50)^{x}(1-0.50)^{10-x}$

(a)

If both the computer and the human wins five games each then the probability of the human winning 5 games is:

$P (X=5)={10\choose 5}(0.50)^{5}(1-0.50)^{10-5}\\=252\times 0.03125\times0.03125\\=0.246094\\\approx 0.2461$

Thus, the probability that each of them win five games is 0.2461.

(b)

If the computer wins 7 games then the number of games won by the human is, 10 - 7 = 3.

P (Computer winning 7 games) = P (Human winning 3 games)

The probability that the human wins 3 games is:

$P (X=3)={10\choose 3}(0.50)^{3}(1-0.50)^{10-3}\\=120\times 0.125\times0.0078125\\=0.1171875\\\approx 0.1172$

Thus, the probability that the computer wins seven games is 0.1172.

(c)

Compute the probability that the human wins at least 7 games as follows:

$P(X\geq 7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)\\={10\choose 7}(0.50)^{7}(1-0.50)^{10-7}+{10\choose 8}(0.50)^{8}(1-0.50)^{10-8}\\+{10\choose 9}(0.50)^{9}(1-0.50)^{10-9}+{10\choose 10}(0.50)^{10}(1-0.50)^{10-10}\\=0.1172+0.044+0.009+0.0009\\=0.1711$

Thus, the probability that the human wins at least 7 games is 0.1711.