Binomial Distribution

[istar]

Out if 56 coins, 18 are rare. If you select 10 of the coins, what us the probability that all of them are rare ?

I found this answer but not sure.

P(all coins) = 18/56 = 0.3214 =0.32 , q= 1-p = 1-0.32=0.68 , n =10
a) All of them (10 trials) are rare:
ncr= 10c10= n!/(r!*(n-r) = 10!/(10-10)!(10)!= ( 10!)/(0)!*(10)!= (10)!/1*(10)! = 1

P(10 trials) are rare) = ncr*(p^r)*(q)^r = 1*(0.32^10)*(0.68)^10
= 1*(1.13)*(0.02)=0.0226

 

[pka]

I would do this way. SEE HERE

 

[Steven G]

1st of all 18/56 does not equal .3216 and .3216 does not equal .32 Why are you rounding?

When you wrote P(all coins) what is that supposed to mean? That is exactly what are you trying to find the probability of?

I suspect that you meant to write that P(of picking a rare coin) = 18/56 = 9/28

The total number of ways to choose 10 coins from 56 is .....

The total number of ways to choose 10 coins from 18 is ...

The total number of ways to choose 0 coins from 38 is ...

 

[istar]

Hi Jomo, you are right . I was referring P(of picking a rare coin). But if I want to get all rare coins out of 10 trials = ncr*(p^r)*(q)^r 10C10(9/28)^10x(1-9/28 )^10

 

[pka]

Hi Jomo, you are right . I was referring P(of picking a rare coin). But if I want to get all rare coins out of 10 trials = ncr*(p^r)*(q)^r 10C10(9/28)^10x(1-9/28 )^10

@istar, This is not classic Binomial Distribution you should read that link.
Unless you have sampling with replacement we do not have independent events.
You can never find all the fake coins.

 

[istar]

Thx pka, my thinking was out 18 rare coins, by selecting 10 out of that coins, the probability of getting exactly all of them rare
ones are for binomial distribution . I guess I am wrong then.

This is what I though was right after couple hrs of work.

ncr= 10c10= n!/(r!*(n-r) = 10!/(10-10)!(10)!= ( 10!)/(0)!*(10)!= (10)!/1*(10)! = 1
P(10 trials are rare) = ncr*(p^r)*(q)^n-r
= 1*(0.321428571)^10*(0.678571429)^10-10
= 1*(0. 321428571)^10*(0.678571429)^0
= 1*(0.0001177185)*(1) = 0.0001177185
Probability that all of them are rare = 0.0001177185

 

[pka]

ncr= 10c10= n!/(r!*(n-r) = 10!/(10-10)!(10)!= ( 10!)/(0)!*(10)!= (10)!/1*(10)! = 1
P(10 trials are rare) = ncr*(p^r)*(q)^n-r Probability that all of them are rare = 0.0001177185

That is not the answer we expect. LOOK HERE
Let's walk through the solution.
There is a collection of fifty-six coins, eighteen of which are very rare. We blindly select ten at random from the well mixed collection.
There are \(\mathcal{C}_{10}^{56}=35607051480\) ways to select ten from a collection of fifty-six.
There are \(\mathcal{C}_{10}^{18}=43758\) ways to select ten from the collection of eighteen rare coins.
If we divide those two we get the probability that all ten are rare.

 

[Steven G]

Look at a video on hyper-geometic method.