Suppose N has a geometric distribution with parameter p. Derive a closed-form expression for E(N | N <= k), k = 1,2,... Check via simulation for p = 0.2, k = 3.

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 ( 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)^yk= 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

if x decreases by 4 then y increases by 10

step-by-step explanation:

answer: 10 meter

step-by-step explanation:

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