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I got this Poisson distribution problem.

There's a clothes shop, and average number of customers per 1 hour are 5 men , 10 women, independent each other, in Poisson distribution. If for 30 minutes there had been totally 10 customers, what's the probability of women customer more than 2?

  • This is how I solved. Poisson average of man customers for 30 minute = m = 2.5

women = w = 5

all = a = 7.5

then the solution is p(w>2 | a=10)

=1- [ p(w>2 ∩ a=10)/p(a=10)]

= 1- { [p(w=0)*p(m=10) + p(w=1) * p(m=9)]/p(a=10)

(solve with Poisson formula)

=0.9996

I'm not quite sure if this is right, because not sure if p(w=1 ∩ a=10) can be interpreted right into 

p(w=1) * p(m=9)

. Is this approach right? or am I missing something Thank you verymuch

서영빈
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    This approach is right. The event ${w=1,a=10}$ coincides with the event ${w=1,m=9}$ and the last is the intersection of two indpendent events. – NCh Jan 27 '20 at 07:02
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    Here's a tutorial and reference for typesetting math on this site. – joriki Jan 27 '20 at 07:29

1 Answers1

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Your approach is generally correct, but where you have $w\gt2$ the second time it should be $w\le2$ instead, and you're missing the $w=2$ case in the sum below that.

However, a somewhat simpler approach would be: Conditional on the total number of customers, each customer is independently a man with probability $\frac13$ and a woman with probability $\frac23$, so the probability for more than $2$ women is

$$ 1-\sum_{w=0}^2\binom{10}w\left(\frac23\right)^w\left(\frac13\right)^{10-w}=\frac{19616}{19683}\approx0.9966\;. $$

See Finding expectation of a Poisson variable when condition has been made.

joriki
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