disagree with that solution.
prob(rare) = 18/56 = 9/28
Prob( 10 of 10 are rare) = (9/28)^10 = .000011771
what Brandon meant was:
C(10,10) (9/28)^10 (19.28)^0
which is 1*(9/28)^10 * 1 = my answer
Out if 56 coins, 18 are rare. If you select 10 of the coins, what us the probability that all of them are rare ?
P(picking rare coin) = 18/56 = 0.3214 q= 1-p = 1-0.3214= 0.6786 , number of trials = 10
a) All rare coins:
ncr= 10c10= n!/(r!*(n-r) = 10!/(10-10)!(10)!= ( 10!)/(0)!*(10)!= (10)!/1*(10)! = 1
P(all rare coins) = ncr*(p^r)*(q)^r = 1*(0.3214^10)*(0.6786)^10
= 1*(0.000008534)*(0.020708)=0.00000017672
2 answers
P(picking rare coin) = 18/56 = 0.321428571 q= 1-p = 1-0.3214= 0.6786 , number of trials = 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)^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
I guess this is the right answer!
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)^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
I guess this is the right answer!