Question #113013
A geological study indicates that an exploratory oil well should strike oil with probability 0.2.
(a) What is the probability that the first strike comes on the third well drilled? (b) What assumptions did you make to obtain the answer in (a).
26. At a checkout counter customers arrive at an average of 2 per minute. Find the probability that
1
Expert's answer
2020-04-30T18:34:42-0400

Let XX be the number of wells drilled until the first first strike of oil.

It is reasonable to assume that XX has a negative binomial distribution.


nb(x;r,p)=(x+r1r1)pr(1p)x,x=0,1,2,...nb(x; r, p) =\binom{x+r-1}{r-1}p^r (1-p)^x,x =0, 1, 2, . . .

Given p=0.2,r=1,x=31=2p=0.2, r=1, x=3-1=2

(a) The probability that the first strike comes on the third well drilled is


(2+1111)0.21(10.2)2=0.128\binom{2+1-1}{1-1}0.2^1(1-0.2)^2=0.128

(b)

1. The experiment consists of a sequence of independent trials.

2. Each trial can result in either a success (S) or a failure (F).

3. The probability of success is constant from trial to trial, so P(S on trial i)=pP(S \ on \ trial \ i)=p for i=1,2,...i=1,2,...

4. The experiment continues (trials are performed) until a total of rr successes have been observed, where rr is a specified positive integer.


26. Assume that customers arrival has Poisson distribution with λ=2.\lambda=2.

Let Pk=P_k= the number of arrivals during the given interval.

Let λt\lambda_t be arrival rate during t minutes, λt=2t\lambda_t=2\cdot t

(a) at most 4 will arrive at any given time


P(k4)=P(k=0)+P(k=1)+P(k=2)+P(k\leq4)=P(k=0)+P(k=1)+P(k=2)+


+P(k=3)+P(k=4)=k=04(2t)ke2tk!+P(k=3)+P(k=4)=\displaystyle\sum_{k=0}^4{(2t)^k e^{-2t}\over k!}

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