Probability when no overlapping condition exists???

[rsgrns]

There are 1000 boxes each containing some object in it. Now, you know that there are 5 boxes amongst those 1000 boxes that contain watch. You need to pick up 20 boxes from 1000 boxes. What is the probability that out of 20 boxes picked from 1000 boxes, exactly one box will be the box containing watch?

Note: Similar to watch, out of 1000 boxes, there could be say 7 boxes containing balls, 10 boxes containing picture frames etc. But we dont know about such cases & we only know that there are 5 boxes containing watches.

Can anyone help me in guiding how to solve this problem?

 

[galactus]

This appears to be a hypergeometric distribution.

boxes and watches.

population, N=1000

sample size, n = 20

k successes = 5

N-k failures = 995

x successes in sample = 1

n-x failures in sample = 19


\(\displaystyle \L\\\frac{C(k,x)C(N-k,n-x)}{C(N,n)}\)

Give it a go. What do you get?.

 

[rsgrns]

Thanks a lot Galactus! Actually, I was doing some research and I got stuck up with this problem. The answer you mentioned seems to be exactly in line with my expectations. In fact, I was trying so many different ways of trying to solve this problem for the past 2 days but nothing materialized.

Just one short question, in the same problem, if I need to know the probability of drawing at least one box that contains watch, then I would calculate it as

P(x>=1) = P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5), with value of N=1000,n=20,k=5.


I am planning to evaluate using this formula in multiple cases. Hope it works & I am able to complete my research successfully! Thanks once again for your prompt & sensible reply!

 

[galactus]

The best way to find 'at least one' is to find the probability of none and subtract from 1. That's because at least one is the opposite of none.

So, in the same problem, find the probability of no watches in a 20 box sample.

\(\displaystyle \L\\\frac{C(5,0)C(995,20)}{C(1000,20)}\approx{0.904}\)

Now, subtract from 1 and get 1-0.904=0.096

The probability of at least one is about 9.6%

 

[soroban]

Hello, rsgrns!

There are 1000 boxes each containing some object in it.
Now, you know that there are 5 boxes among those 1000 boxes that contain a watch.
You need to pick up 20 boxes from 1000 boxes.
What is the probability that out of 20 boxes, exactly one box will contain a watch?

There are 5 boxes with Watches and 995 boxes with some Other object.

There are: \(\displaystyle \:{1000\choose20}\) possible ways to pick 20 boxes.

To get one Watch and nineteen Others, there are: \(\displaystyle \:{5\choose1}{995\choose19}\) ways.

Therefore, the probability is: \(\displaystyle \L\:\frac{{5\choose1}{995\choose19}}{{1000\choose20}}\)

[This is the result from Galactus' formula.]