Answer:
P(X= k) = (1-p)^k-1.p
Step-by-step explanation:
Given that the number of trials is
N < = k, the geometric distribution gives the probability that there are k-1 trials that result in failure(F) before the success(S) at the kth trials.
Given p = success,
1 - p = failure
Hence the distribution is described as: Pr ( FFFF.....FS)
Pr(X= k) = (1-p)(1-p)(1-p)....(1-p)p
Pr((X=k) = (1 - p)^ (k-1) .p
Since N<=k
Pr (X =k) = p(1-p)^k-1, k= 1,2,...k
0, elsewhere
If the probability is defined for Y, the number of failure before a success
Pr (Y= k) = p(1-p)^y......k= 0,1,2,3
0, elsewhere.
Given p= 0.2, k= 3,
P(X= 3) =( 0.2) × (1 - 0.2)²
P(X=3) = 0.128
Answer: c
Step-by-step explanation:
