Have I done this binomial distribution question correctly?

#1
Insurance policies are sold to 10 different people age 25-30 years. All of these are in good health
conditions. The probability that a person of similar condition will enjoy more than 25 years is 4/5. Calculate
the probability that in 25 years
a) Almost 2 will die.

It will be P(X=0) +P(X=1)+P(X=2)

probability of living = 4/5 = 0.8
prob of dying = 0.2

So p will be prob of dying = 0.2 and hence q will be 0.8

So P(X=0) : 10C0*(prob of dying)^x *(prob of not dying)^n-x
= 10C0*(prob of dying)^x *(prob of not dying)^n-x
= 10C0*(0.2)^0 *(0.8)^10-0
= 0.10737
So P(X=1) : 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C1*(0.2)^1 *(0.8)^10-1
= 0.268435
So P(X=2) : 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C2*(0.2)^2 *(0.8)^10-2
= 0.30198

So sum them all = 0.6777 Ans

Please tell have I done it correctly?
 

obh

Well-Known Member
#2
Insurance policies are sold to 10 different people age 25-30 years. All of these are in good health
conditions. The probability that a person of similar condition will enjoy more than 25 years is 4/5. Calculate
the probability that in 25 years
a) Almost 2 will die.

It will be P(X=0) +P(X=1)+P(X=2)

probability of living = 4/5 = 0.8
prob of dying = 0.2

So p will be prob of dying = 0.2 and hence q will be 0.8

So P(X=0) : 10C0*(prob of dying)^x *(prob of not dying)^n-x
= 10C0*(prob of dying)^x *(prob of not dying)^n-x
= 10C0*(0.2)^0 *(0.8)^10-0
= 0.10737
So P(X=1) : 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C1*(0.2)^1 *(0.8)^10-1
= 0.268435
So P(X=2) : 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C1*(prob of dying)^x *(prob of not dying)^n-x
= 10C2*(0.2)^2 *(0.8)^10-2
= 0.30198

So sum them all = 0.6777 Ans

Please tell have I done it correctly?
Hi Sallu,

I assume you mean at most (not almost), hence p(X2)
If so you got the correct answer.
You may check your answer in the following calculator: https://www.statskingdom.com/1_binomial_distribution.html