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
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