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Wize University Statistics Textbook > Discrete Probability Distributions

Approximating the Binomial Distribution with Poisson

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Approximating the Binomial Distribution with Poisson

Suppose that XX is binomial with nn trials and a probability of success pp.

We can approximate the binomial distribution with a Poisson distribution if the following conditions hold:

  1. np<5np<5 or n(1−p)<5n(1-p)<5 (some textbooks say 10)
  2. nn is very large (most textbooks say 20)

Then, the Poisson mean is approximately equal to the Binomial mean:

λ=np\lambda=np

And the formula becomes:
P(x)=λxe−λx!=(np)xe−npx!\boxed{P\left(x\right)=\frac{\lambda^xe^{-\lambda}}{x!}=\frac{\left(np\right)^xe^{-np}}{x!}}

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Example: Approximating Binomial with Poisson

A majestic bird can lay many, many eggs to feed the village. Suppose there are two possible outcomes for each egg: shiny or not shiny. (Shiny eggs make great frittatas!) We assume 100 eggs are laid in an identical and independent manner, and the probability of getting a shiny egg is constant at 4%.


(a) Can we approximate the Binomial distribution with Poisson?

Yes, because nq=0.04(100)=4<5nq=0.04\left(100\right)=4<5.


(b) What is the probability of getting at least 3 shiny eggs? [Use the binomial formula]

P(X⩾3)P\left(X\geqslant3\right)
=1−P(X⩽2)=1-P\left(X\leqslant2\right)
=1−P(X=2)−P(X=1)−P(X=0) =1-P\left(X=2\right)-P\left(X=1\right)-P\left(X=0\right)\

Using the binomial formula:
=1−[(100C2)(0.04)2(0.96)98]−[(100C1)(0.04)1(0.96)99]−[(100C0)(0.04)0(0.96)100]=1-\left[\left(_{100}C_{2}\right)\left(0.04\right)^2\left(0.96\right)^{98}\right]-\left[\left(_{100}C_{1}\right)\left(0.04\right)^1\left(0.96\right)^{99}\right]-\left[\left(_{100}C_{0}\right)\left(0.04\right)^0\left(0.96\right)^{100}\right]
≈1−0.14498−0.07029−0.01687\approx1-0.14498-0.07029-0.01687
≈0.76786\approx0.76786



(c) What is the probability of getting at least 3 shiny eggs? [Use the Poisson formula]


λ=4\lambda=4

P(X⩾3)P\left(X\geqslant3\right)
=1−P(X⩽2)=1-P\left(X\leqslant2\right)
=1−P(X=2)−P(X=1)+P(X=0) =1-P\left(X=2\right)-P\left(X=1\right)+P\left(X=0\right)\

Using the binomial formula:
=1−e−4(4)22!−e−4(4)11!−e−4(4)00!\displaystyle =1-\frac{e^{-4}\left(4\right)^2}{2!}-\frac{e^{-4}\left(4\right)^1}{1!}-\frac{e^{-4}\left(4\right)^0}{0!}
=1−0.14653−0.07326−0.01832=1-0.14653-0.07326-0.01832
=0.76189=0.76189

Very close answers!


Practice: Approximating Binomial with Poisson

Five percent of customers will make a refund. If there were 60 independent customers, what is the probability that 4 customers will make a refund?