Question

we have an urn with 10 balls numbered from 1 to 10. we choose a sample of 111 with replacement. approximate the probability of the event that the number one appears at most 3 times in the sample. use both the normal and the poisson approximation, and compare the results with the exact probability 0.00327556.

Answer 1

The exact probability is less than probability using Poisson distribution is less than probability using normal distribution.

Given that,

10 balls, numbered from 1 to 10, are in an urn. We selected a 111 person sample with replacement.

We have to calculate the probability that the number one appears in the sample no more than three times. Compare the outcomes with the precise probability using both the normal and the Poisson approximations 0.00327556.

We know that,

Let

X = number of times that 1 appears in a sample

Probability of getting 1 =1/10=0.1

Then X congruent to Binomial (111,0.1)

Now,

If we approximately X is Poisson distribution with parameter

λ = 111 × 0.1 = 11.1

Then

P(X≤3)=∑x=0 to 3 e⁻¹¹°¹(11.1)ˣ/x!

P(X≤3)= 0.00327556

If we approximate X by Normal distribution with mean

μ = 111 × 0.1 = 11.1

σ = √111 × 0.1 × 0.9 = 3.160696 then

P(X≤3) = 0.005192688

Exact probability is

P(X≤3) = 0.003275558

Hence the answer is, The exact probability is less than probability using Poisson distribution is less than probability using normal distribution.

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