Question

A box of 12 tins of tuna contains 6 which are tainted, suppose 7 tins are opened for inspection, find the probability that (a) exactly 3 of them are tainted, (b) at least 5 of them are tainted, (c) at most 3 of them are tainted

Answer 1

Let $X=$ the number of tainted tins: $X\sim Bin(n, p)$

Given,

 $n=7,\\ p=6/12=0.5, \\q=1-p=0.5$

(a)  exactly 3 of them are tainted,

$$P(X=3)=\dbinom{7}{3}(0.5)^3(0.5)^{7-3}=0.2734375$$

(b) at least 5 of them are tainted,

$$P(X\geq 5)=P(X=5)+P(X=6)+P(X=7)$$ $$=\dbinom{7}{5}(0.5)^5(0.5)^{7-5}+\dbinom{7}{6}(0.5)^6(0.5)^{7-6}$$ $$+\dbinom{7}{7}(0.5)^7(0.5)^{7-7}=0.2265625$$

(c)  at most 3 of them are tainted

$$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)$$ $$+P(X=3)=\dbinom{7}{0}(0.5)^0(0.5)^{7-0}$$ $$+\dbinom{7}{1}(0.5)^1(0.5)^{7-1}+\dbinom{7}{2}(0.5)^2(0.5)^{7-2}$$ $$+\dbinom{7}{3}(0.5)^3(0.5)^{7-3}=0.5$$