Brock Brown
2021-12-17
Answered

An oil drilling company ventures into various locations, and its success or failure is independent from one location to another. Suppose the probability of a success at any specific location is 0.30. What is the probability that a driller drills 10 locations and finds 1 success? The driller feels that he will go bankrupt if he drills 10 times before the first success occurs. What are the driller's prospects for bankruptcy? i. ii.

You can still ask an expert for help

Kirsten Davis

Answered 2021-12-18
Author has **27** answers

Step 1

Given Information:

Probability of success at any specific location i.e.,$p=0.30$

i) To find the probability that a driller drills 10 locations and finds 1 success:

Let X be a random variable which denotes the number of successes among 10 drills. Because trials are independent, X follows a Binomial distribution with parameters$n=10\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}p=0.30$ .

Probability mass function of Binomial variable X is given as:

$P(X=x){=}^{n}{C}_{x}{p}^{x}{(1-p)}^{n-x}$

Required probability is obtained as follows:

$P(X=1){=}^{10}{C}_{1}{\left(0.30\right)}^{1}{(1-0.30)}^{10-1}$

$=\frac{10!}{1!(10-1)!}\times 0.30\times {\left(0.70\right)}^{9}$

$=\frac{10\times 9!}{9!}\times 0.30\times 0.040353607$

$=10\times 0.0121060821$

$=0.121060821$

$\approx 0.1211$

Thus, the probability that a driller drills 10 locations and finds 1 success is 0.1211

Step 2

ii) Let X represent the number of drills required to get first success.

Find the probability that the driller prospects for bankruptcy.

In other words, that there are no successes$(X=0)$ in 10 trials:

Required probability is obtained as follows:

$P(X=0){=}^{10}{C}_{0}{\left(0.30\right)}^{0}{(1-0.30)}^{10-0}$

$=\frac{10!}{0!(10-0)!}\times 1\times {\left(0.70\right)}^{10}$

$=\frac{10!}{10!}\times 1\times 0.0282475249$

$=0.0282475249$

$\approx 0.0282$

Given Information:

Probability of success at any specific location i.e.,

i) To find the probability that a driller drills 10 locations and finds 1 success:

Let X be a random variable which denotes the number of successes among 10 drills. Because trials are independent, X follows a Binomial distribution with parameters

Probability mass function of Binomial variable X is given as:

Required probability is obtained as follows:

Thus, the probability that a driller drills 10 locations and finds 1 success is 0.1211

Step 2

ii) Let X represent the number of drills required to get first success.

Find the probability that the driller prospects for bankruptcy.

In other words, that there are no successes

Required probability is obtained as follows:

godsrvnt0706

Answered 2021-12-19
Author has **31** answers

Suppose the probability of a success at any specific location is 0.25.

i) What is the probability that a driller drills 10 locations and finds 1 success?

$P\left(\text{one success in 10}\right)=\left(10C1\right){\left(0.25\right)}^{1}{\left(0.75\right)}^{9}=0.1877$ ...

ii) The driller feels that he will go bankrupt if he drills 10 times before the first success occurs. What are the driller is prospects for bankruptcy?

$P\left(\text{no success in 10 trials}\right)={0.75}^{10}=0.0563$ ...

Chances are 563 in 10000 that he will go bankrupt.

i) What is the probability that a driller drills 10 locations and finds 1 success?

ii) The driller feels that he will go bankrupt if he drills 10 times before the first success occurs. What are the driller is prospects for bankruptcy?

Chances are 563 in 10000 that he will go bankrupt.

nick1337

Answered 2021-12-28
Author has **510** answers

Step 1

- Consider that drilling in a location is a trial.

- Probability of success in each trial is p=0.25,

- Probability of failure is q=1-p=1-0.25=0.75

Step 2

a) Let random variable X represent the number of success among 10 drills.

Because trials are independent, X has binomial distribution with parameters n=10 and p=0.25.

Probability mass function of X is

Lets calculate the probability the driller drills at at 10 locations and has 1 success.

=0.1877

Step 3

b) Let random variable X represent the number of drills required to get first success.

Now, lets find the probability that the driller prospects for bankruptcy, in other words, that there are no successes in 10 trials.

=0.05631

asked 2021-05-21

At a certain college, 6% of all students come from outside the United States. Incoming students there are assigned at random to freshman dorms, where students live in residential clusters of 40 freshmen sharing a common lounge area. How many international students would you expect to find in a typical cluster? With what standard deviation?

asked 2021-09-24

If X has a binomial distribution, $n=5\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}p=0.4$ . Determine the probabilities of $X=0$ .

a). -1.2321

b) 0.07776

c) 0.1073741824

d) 0.1534621

a). -1.2321

b) 0.07776

c) 0.1073741824

d) 0.1534621

asked 2021-09-25

In Exercises, X denotes a binomial random variable with parameters n and p. For each exercise, indicate whicharea under the appropriate normal curve would be determined to approximate the specified binomial
probability.

$P(7\le X\le 10)$

asked 2021-09-30

X denotes a binomial random variable with parameters n and p. For each exercise, indicate which area under the appropriate normal curve would be determined to approximate the specified binomial probability.

$P(X\le 4)$

asked 2021-09-21

Consider a binomial experiment with 15 trials and probability 0.45 of success on a single trial.

(a) Use the binomial distribution to find the probability of exactly 10 successes. (Round your answer to three decimal places.)

(b) Use the normal distribution to approximate the probability of exactly 10 successes. (Round your answer to three decimal places.)

(a) Use the binomial distribution to find the probability of exactly 10 successes. (Round your answer to three decimal places.)

(b) Use the normal distribution to approximate the probability of exactly 10 successes. (Round your answer to three decimal places.)

asked 2021-02-19

Convert the binomial probability to a normal distribution probability using continuity correction.
P (x = 45).

asked 2021-12-17

Tell, what is the factorial of 10?