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

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)

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

$=0.060607$