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Jan 8, 2014
 #2
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Jan 8, 2014
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Jan 8, 2014
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Sorry in advance for my english, it's not perfect and I might make some grammar mistake
First, i forgot to mention that the result should be [0.270842].

You are d**n right I did the right reasoning, but I made ​​some calculation errors (shame on me, you do not study mathematics at night!)
To apologize i'll post the solved exercize so it's avaible to other users.

Since they are repeated tests, you will have to apply Bernoulli:
P=nCr(n.k)*p^k*(1-p)^(n-k)
where, in this case, n=25 b***s, p=16/25 (possibilities to pick a white ball) and k is the number of the white b***s i am supposed to pick.

There are two ways to solve this:
1) To add P(0), P(1), P(2), P(3), P(4), P(5)
2) P=1-P(6)-P(7)-P(8)-P(9)-P(10)

Let's pick the first one. We need to apply Bernoulli to every case (no white b***s picked, 1 white ball picked, 2 white ball picked etc.)
P(0)= nCr(10,0)*(16/25)^0*(1-16/25)^10 = 0.0000365615844006
P(1) = nCr(10,1)*(16/25)^1*(1-16/25)^9 = 0.0006499837226779
P(2) = nCr(10,2)*(16/25)^2*(1-16/25)^8 = 0.0051998697814229
P(3)= nCr(10,3)*(16/25)^3*(1-16/25)^7 = 0.0246512345193382
P(4)= nCr(10,4)*(16/25)^4*(1-16/25)^6 = 0.0766927296157188
P(5) = nCr(10,5)*(16/25)^5*(1-16/25)^5 = 0.1636111565135335

We need to keep in mind that the events are incompatibles (if i pick 1 ball, i con't pick2 at the same time) and unconditioned (if i pick a ball and i put it again in the urn, the ball number will always be the same). The probability that less than 6 b***s are white is obtained adding the calculated probabilities between them:
P= 0.0000365615844006+0.0006499837226779+0.0051998697814229+0.0246512345193382+0.0766927296157188+0.1636111565135335 = 0.2708415357370919 = 0.270842

And here you go thanks melody
Jan 8, 2014

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