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on the way to work a guy must pass through $10$ traffic lights suppose that in the long run he encounters a red light at $40$% of these signals and whether any particular signal is red is independent of whether any other one is red

On what proprotion of days will our friend encounter at most two red lights?

My working
$40$% of 10 is 4, so I take lambda = 4

"At most two" means 0 red lights + 1 red light + 2 red lights.

So this must mean that we can use a poisson cdf function with lambda = 4 and $x$ = 2 which in the poisson distribution table is 0.2381

However, the correct answer is actually $0.167$

What am I missing?

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When $n$ is large and $np$ is small, (say $n=200$, $np\approx 1$) then the Poisson distribution can give a useful approximation to the binomial, but $n=10$ is far too small to give a decent approximation when $np=4$. – André Nicolas Oct 25 '11 at 14:14
up vote 2 down vote accepted

The number of red lights he hits is the sum of 10 indicator random variables, that are 1 with probability $p=0.4$ and 0 with probability $q=0.6$. These are known as Bernoulli random variables. The sum of $n$ Bernoulli random variables follows binomial distribution with $\mathbb{P}(X=k) = \binom{n}{k} p^k q^{n-k}$.

The quantity you seek is: $$ \mathbb{P}(X=0) + \mathbb{P}(X=1) + \mathbb{P}(X=2) \stackrel{n=10}{=} (-1 + p)^8 (1 + 8 p + 36 p^2) = 0.16729 $$

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Ahh.. thanks, I somehow got the idea that this was specifically a poisson distribution question and so I went ahead and assumed I had to use poisson. But I see now that binomial was what I needed. thanks again – Arvin Oct 25 '11 at 13:03

Try to calculate the probability using a binomial distribution. This is not a case where the Poisson distribution is a good fit.

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