Answer to Question #208417 in Statistics and Probability for BUTT

Question

In a certain district the covid patients recover by vaccine is 0.4. If 10 people are infected by this
disease. What is the probability ,
i. At least 6 survive.
ii. From 2 to 7 survive.
iii. Exactly 3 survive.
iv. Exactly 2 are not surviving.
v. Also find the expected value.

Answer 1

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

Given $p=0.4, n=10.$

$q=1-p=1-0.4=0.6$

i.

$$P(X\geq6)=P(X=6)+P(X=7)+P(X=8)$$

$$+P(X=9)+P(X=10)$$

$$=\dbinom{10}{6}(0.4)^6(0.6)^{10-6}+\dbinom{10}{7}(0.4)^6(0.6)^{10-7}$$

$$+\dbinom{10}{8}(0.4)^8(0.6)^{10-8}+\dbinom{10}{9}(0.4)^9(0.6)^{10-9}$$

$$+\dbinom{10}{10}(0.4)^{10}(0.6)^{10-10}=0.1662386176$$

ii.

$$P(2\leq X\leq7)=P(X=2)+P(X=3)$$

$$+P(X=4)+P(X=5)+P(X=6)+P(X=7)$$

$$=\dbinom{10}{2}(0.4)^2(0.6)^{10-2}+\dbinom{10}{3}(0.4)^3(0.6)^{10-3}$$

$$+\dbinom{10}{4}(0.4)^4(0.6)^{10-4}+\dbinom{10}{5}(0.4)^5(0.6)^{10-5}$$

$$+\dbinom{10}{6}(0.4)^{6}(0.6)^{10-6}+\dbinom{10}{7}(0.4)^{7}(0.6)^{10-7}$$ $$= 0.9413480448$$

iii.

$$P(X=3)=\dbinom{10}{3}(0.4)^3(0.6)^{10-3}$$

$$=0.214990848$$

iv.

$$P(X=10-2)=\dbinom{10}{8}(0.4)^8(0.6)^{10-8}$$

$$=0.010616832$$

v.

$$E(X)=np=10(0.4)=4$$