Answer to Question #208417 in Statistics and Probability for BUTT

Question #208417
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.
1
Expert's answer
2021-06-21T07:38:25-0400

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

Given p=0.4,n=10.p=0.4, n=10.

q=1p=10.4=0.6q=1-p=1-0.4=0.6

i.

P(X6)=P(X=6)+P(X=7)+P(X=8)P(X\geq6)=P(X=6)+P(X=7)+P(X=8)

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

=(106)(0.4)6(0.6)106+(107)(0.4)6(0.6)107=\dbinom{10}{6}(0.4)^6(0.6)^{10-6}+\dbinom{10}{7}(0.4)^6(0.6)^{10-7}

+(108)(0.4)8(0.6)108+(109)(0.4)9(0.6)109+\dbinom{10}{8}(0.4)^8(0.6)^{10-8}+\dbinom{10}{9}(0.4)^9(0.6)^{10-9}

+(1010)(0.4)10(0.6)1010=0.1662386176+\dbinom{10}{10}(0.4)^{10}(0.6)^{10-10}=0.1662386176

ii.

P(2X7)=P(X=2)+P(X=3)P(2\leq X\leq7)=P(X=2)+P(X=3)

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

=(102)(0.4)2(0.6)102+(103)(0.4)3(0.6)103=\dbinom{10}{2}(0.4)^2(0.6)^{10-2}+\dbinom{10}{3}(0.4)^3(0.6)^{10-3}

+(104)(0.4)4(0.6)104+(105)(0.4)5(0.6)105+\dbinom{10}{4}(0.4)^4(0.6)^{10-4}+\dbinom{10}{5}(0.4)^5(0.6)^{10-5}

+(106)(0.4)6(0.6)106+(107)(0.4)7(0.6)107+\dbinom{10}{6}(0.4)^{6}(0.6)^{10-6}+\dbinom{10}{7}(0.4)^{7}(0.6)^{10-7}=0.9413480448= 0.9413480448

iii.


P(X=3)=(103)(0.4)3(0.6)103P(X=3)=\dbinom{10}{3}(0.4)^3(0.6)^{10-3}

=0.214990848=0.214990848

iv.


P(X=102)=(108)(0.4)8(0.6)108P(X=10-2)=\dbinom{10}{8}(0.4)^8(0.6)^{10-8}

=0.010616832=0.010616832

v.


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


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