Question

In a challenging chess game, the probability that Mr.B will win is 0.67. Assume that the chess competition follows a binomial distribution and 7 separate matches will be played. 

Determine: 

(a) will win at most 3 matches 

(b) will lose exactly 5 matches 

(c) will win every match

Answer 1

Let $X=$ the number of matches won by Mr.B: $X\sim Bin(n, p).$

Given $n=7, p=0.67, q=1-p=1-0.67=0.33$

(a)

$$P(X\le 3)=P(X=0)+P(X=1)$$

$$+P(X=2)+P(X=3)$$

$$=\dbinom{7}{0}(0.67)^0(0.33)^{7-0}+\dbinom{7}{1}(0.67)^1(0.33)^{7-1}$$

$$+\dbinom{7}{2}(0.67)^2(0.33)^{7-2}+\dbinom{7}{3}(0.67)^3(0.33)^{7-3}$$

$$=0.00042618443+0.00605698477$$

$$+0.03689254363+0.1248384052$$

$$=0.168214$$

(b)

$$7-5=2$$

$$P(X=2)=\dbinom{7}{2}(0.67)^2(0.33)^{7-2}$$

$$=0.036893$$

(c)

$$P(X=7)=\dbinom{7}{7}(0.67)^7(0.33)^{7-7}$$

$$=0.060607$$