/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 An investor has the option of in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An investor has the option of investing in three of five recommended stocks. Unknown to her, only two will show a substantial profit within the next 5 years. If she selects the three stocks at random (giving every combination of three stocks an equal chance of selection), what is the probability that she selects the two profitable stocks? What is the probability that she selects only one of the two profitable stocks?

Short Answer

Expert verified
Answer: The probability that the investor selects the two profitable stocks is 0.3, and the probability that she selects only one of the two profitable stocks is 0.6.

Step by step solution

01

Calculate the total number of ways the investor can select three stocks

From the problem, the investor has to choose three stocks from a total of five stocks. To calculate the total number of possible combinations, we can use the combination formula. C(n, k) = n! / (k!(n-k)!) So, the total number of ways to select three stocks from the given five stocks is: C(5, 3) = 5! / (3!(5-3)!) = 10
02

Calculate the number of ways the investor can select the two profitable stocks

Since there are two profitable stocks among the five recommended stocks, we want to know the number of ways the investor can select those two stocks. The third stock selected must be one of the three remaining non-profitable stocks. The number of ways to select the two profitable stocks is: C(2, 2) = 2! / (2!(2-2)!) = 1 The number of ways to select the third non-profitable stock is: C(3, 1) = 3! / (1!(3-1)!) = 3 Now, we multiply both results to obtain the total number of ways the investor can select the two profitable stocks and one non-profitable stock: 1 * 3 = 3
03

Calculate the probability of selecting the two profitable stocks

To calculate the probability that the investor selects the two profitable stocks, we divide the number of ways the investor can select the two profitable stocks (calculated in Step 2) by the total number of ways to select three stocks from the given five stocks (calculated in Step 1). Probability of selecting the two profitable stocks = 3 (number of ways to select two profitable stocks) / 10 (total number of ways to select three stocks) = 3/10 = 0.3
04

Calculate the number of ways the investor can select only one profitable stock

Now we want to find the number of ways the investor can select only one of the two profitable stocks. To do this, we calculate the number of ways to select one profitable stock from the two, which is: C(2, 1) = 2! / (1!(2-1)!) = 2 Then, we need to calculate the number of ways to select the remaining two non-profitable stocks from the three non-profitable stocks: C(3, 2) = 3! / (2!(3-2)!) = 3 Now, we multiply both results to obtain the total number of ways the investor can select one profitable stock and two non-profitable stocks: 2 * 3 = 6
05

Calculate the probability of selecting only one profitable stock

To calculate the probability that the investor selects only one of the two profitable stocks, we divide the number of ways the investor can select only one profitable stock (calculated in Step 4) by the total number of ways to select three stocks from the given five stocks (calculated in Step 1). Probability of selecting only one profitable stock = 6 (number of ways to select one profitable stock) / 10 (total number of ways to select three stocks) = 6/10 = 0.6 So the probability that the investor selects the two profitable stocks is 0.3, and the probability that she selects only one of the two profitable stocks is 0.6.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Refer to Exercise 33. Suppose you are interested in following two independent traits in snap peas-seed texture \((\mathrm{S}=\) smooth \(, \mathrm{s}=\) wrinkled \()\) and seed color \((\mathrm{Y}=\) yellow, \(\mathrm{y}=\) green \()-\) in a secondgeneration cross of heterozygous parents. Remember that the capital letter represents the dominant trait. Complete the table with the gene pairs for both traits. All possible pairings are equally likely. a. What proportion of the offspring from this cross will have smooth yellow peas? b. What proportion of the offspring will have smooth green peas? c. What proportion of the offspring will have wrinkled yellow peas? d. What proportion of the offspring will have wrinkled green peas? e. Given that an offspring has smooth yellow peas, what is the probability that this offspring carries one s allele? One s allele and one y allele?

A group of research proposals was evaluated by a panel of experts to decide whether or not they were worthy of funding. When these same proposals were submitted to a second independent panel of experts, the decision to fund was reversed in \(30 \%\) of the cases. If the probability that a proposal is judged worthy of funding by the first panel is \(.2,\) what are the probabilities that: a. A worthy proposal is approved by both panels. b. A worthy proposal is disapproved by both panels. c. A worthy proposal is approved by one panel.

Different illnesses can produce identical symptoms. Suppose a particular set of symptoms, which we will denote as event \(H,\) occurs only when any one of three illnesses- \(A, B,\) or \(C-\) occurs. (For the sake of simplicity, we will assume that illnesses \(A, B,\) and \(C\) are mutually exclusive.) Studies show these probabilities of getting the three illnesses: $$ \begin{array}{l} P(A)=.01 \\ P(B)=.005 \\ P(C)=.02 \end{array} $$ The probabilities of developing the symptoms \(H,\) given a specific illness, are $$ \begin{array}{l} P(H \mid A)=.90 \\ P(H \mid B)=.95 \\ P(H \mid C)=.75 \end{array} $$ Assuming that an ill person shows the symptoms \(H\), what is the probability that the person has illness \(A ?\)

Use the mn Rule to find the number. There are two groups of distinctly different items, 10 in the first group and 8 in the second. If you select one item from each group, how many different pairs can you form?

A French restaurant offers a special summer menu in which, for a fixed dinner cost you can choose from one of two salads, one of two entrees, and one of two desserts. How many different dinners are available?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.