Answer to Question #109141 in Statistics and Probability for Syed Zuhad Hamid

Question

The Corona Virus is known to kill its target rarely and known to hit only 0.002 of the targetswhich he attempts. If the virus hit the 1750individuals, then find the probability:
a) It will kill none of the target
b) It will kill at- most 3 targets
c) It will killBetween 5 to 7 targets

Answer 1

Define the random variable $X$ - the number of the target it will kill

$p=0.002, \ n=1750$

Then $X \backsim \text{Binomial}(n,p)$

For Binomial distribution we have: $P(X=k) =\binom{n}{k}p^k(1-p)^{n-k}$

a) $P(X=0)$ - the probability that it will kill none of the target

$P(X=0)=\binom{1750}{0}\times 0.002^0\times 0.998^{1750}\approx 0.03$

b) $P(X\leq 3)$ - the probability that it will kill at most 3 targets 

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)= \binom{1750}{0}\times 0.002^0\times 0.998^{1750}+ \binom{1750}{1}\times 0.002^1\times 0.998^{1749}+ \binom{1750}{2}\times 0.002^2\times 0.998^{1748}+ \binom{1750}{3}\times 0.002^3\times 0.998^{1747}\approx 0.46$

c) $P(5\leq X\leq 7)$ - the probability that it will kill between 5 to 7 targets

$P(5\leq X\leq 7)=P(X=5)+P(X=6)+P(X=7)= \binom{1750}{5}\times 0.002^5\times 0.998^{1745}+ \binom{1750}{6}\times 0.002^6\times 0.998^{1744}+ \binom{1750}{7}\times 0.002^7\times 0.998^{1743}\approx 0.235$

Answer: a) 0.03 b) 0.46 c) 0.235