# Have I done this binomial distribution question correctly?

#### sallu110

##### New Member
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
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.

#### obh

##### Well-Known Member
I know! What should I do then? Will it be P(X=0) and P(X=1) only?
I don't know what is almost... Is 1 almost 2?
The binomial distribution is discrete. So if it isn't include 2 it is only 0 and 1.

#### sallu110

##### New Member
I don't know what is almost... Is 1 almost 2?
The binomial distribution is discrete. So if it isn't include 2 it is only 0 and 1.
Alright! Thanks bro!
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