Answer to Question #113013 in Statistics and Probability for Gideon Danso

Question

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

Answer 1

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

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

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

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

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

$$\binom{2+1-1}{1-1}0.2^1(1-0.2)^2=0.128$$

(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 $P(S \ on \ trial \ i)=p$ for $i=1,2,...$

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

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

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

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

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

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

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