This is NOT a binomial distribution because the population mean of defects is not known.
Use hypergeometric distribution instead, since the number of defects in the sample (batch) is known, and samples are taken without replacement.
Let
D=number of defective (low resistance) samples in batch=5
d=number of defective samples in draw
N=number of normal (high resistance) samples in batch=15
n=number of defective samples in draw.
Then by the hypergeometric distribution
P(D,d,N,n)=C(D,d)*C(N,n)/C(D+N,d+n)
Since four samples are selected from the twenty, we have d+n=4.
(a) n=1 => d=4-1=3
P(n=1)=C(15,1)*C(5,3)/C(20,4) = 15*10/4845 = 10/323 = 0.03096
(b) n>=1 => d<4-1=3
P(n>=1)
=1-P(n=0)
=1-C(15,0)*C(5,4)/C(20,4)
= 1-1*5/4845
= 1-1/969
= 1- 0.00103
= 0.99897
(c) n>=3
P(n>=3)
=P(n=3)+P(n=4)
= C(15,3)C(5,1)/C(20,4)+C(15,4)C(5,0)/C(20,4)
=455*5/4845+1365*1/4845
=455/969+1365/4845
=0.46956+0.28173
=0.75129
(d) For this part, we need to assume independence of low resistance and scratches.
We proceed separate to find probabilities of 1 sample with low resistance/scratch.
P(d=1) [ one part shows low resistance]
=C(15,3)C(5,1)/C(20,4)
=455*5/4845
=455/969
P(s=1) [ one part shows scratches ]
=C(14,3)C(6,1)/C(20,4)
=364*6/4845
=728/1615
Since the two effects are independent, the probability of both happening is the product of the individual probabilities
=455/969*728/1615
=66248/312987
=0.21166
