Answer to Question #277219 in Statistics and Probability for ayang

Question

The probability that a patient recovers from a rare blood disease is 0.4. If 15 randomly chosen people are known to have contracted this disease, what is the probability that

(a) exactly 8 survive?

(b) at most 2 survive?

(c) less than 2 survive?

(d) at least 13 survive?

(e) more than 13 survive?

(f) between 5 and 8, exclusive, survive?

(g) between 5 and 8, inclusive, survive?

2. From the problem in number 1, how many are expected to survive from the 15 patients?

Answer 1

Let $X=$ the number of patients who survived: $X\sim Bin (n, p).$

Given $n=15, p=0.4, q=1-p=1-0.4=0.6.$

(a)

$$P(X=8)=\dbinom{15}{8}(0.4)^8(0.6)^{15-8}$$

$$=0.11805577445376\approx0.118056$$

(b)

$$P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)$$

$$=\dbinom{15}{0}(0.4)^0(0.6)^{15-0}+\dbinom{15}{1}(0.4)^1(0.6)^{15-1}$$

$$+\dbinom{15}{2}(0.4)^2(0.6)^{15-2}$$

$$=0.02711400077216\approx0.027114$$

(c)

$$P(X< 2)=P(X=0)+P(X=1)$$

$$=\dbinom{15}{0}(0.4)^0(0.6)^{15-0}+\dbinom{15}{1}(0.4)^1(0.6)^{15-1}$$

$$=0.005172034830336\approx0.005172$$

(d)

$$P(X\geq 13)=P(X=13)+P(X=14)+P(X=15)$$

$$=\dbinom{15}{13}(0.4)^{13}(0.6)^{15-13}+\dbinom{15}{14}(0.4)^{14}(0.6)^{15-14}$$

$$+\dbinom{15}{15}(0.4)^{15}(0.6)^{15-15}$$

$$=0.000278904438784\approx0.000279$$

(e)

$$P(X>13)=P(X=14)+P(X=15)$$

$$=\dbinom{15}{14}(0.4)^{14}(0.6)^{15-14}+\dbinom{15}{15}(0.4)^{15}(0.6)^{15-15}$$

$$=0.000025232932864\approx0.000025$$

(f)

$$P(5<X<8)=P(X=6)+P(X=7)$$

$$=\dbinom{15}{6}(0.4)^{6}(0.6)^{15-6}+\dbinom{15}{7}(0.4)^{7}(0.6)^{15-7}$$

$$=0.38368126697472\approx0.383681$$

(g)

$$P(5\leq X\leq 8)=P(X=5)+P(X=6)$$

$$+P(X=7)+P(X=8)=$$

$$=\dbinom{15}{5}(0.4)^{5}(0.6)^{15-5}+\dbinom{15}{6}(0.4)^{6}(0.6)^{15-6}$$

$$=\dbinom{15}{7}(0.4)^{7}(0.6)^{15-7}+\dbinom{15}{8}(0.4)^{8}(0.6)^{15-8}$$

$$=0.687674886193152\approx0.687675$$

2.

$$E(X)=np=15(0.4)=6$$