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A mutual fund company offers its customers a varietyof funds: a money-market fund, three different bond funds (short, intermediate, and long-term), two stock funds (moderate and high-risk), and a balanced fund.

Among customers who own shares in just one fund,the percentages of customers in the different funds areas follows:

Money-market 20% High-risk stock 18%

Short bond 15% Moderate-risk stock 25%

Intermediate bond 10% Balanced 7%

Long bond 5%

A customer who owns shares in just one fund is randomlyselected.

a. What is the probability that the selected individualowns shares in the balanced fund?

b. What is the probability that the individual owns shares in a bond fund?

c. What is the probability that the selected individual does not own shares in a stock fund?

Short Answer

Expert verified

a. The probability that the selected individual owns shares in the balanced fund is 0.07

b. The probability that the selected individual owns shares in a bond fund is 0.30

c. The probability that the selected individual does not owns shares in a stock fund is 0.57

Step by step solution

01

Given information

A mutual fund company offers a variety of funds; they are a money-market fund, three different bond funds (short, intermediate, and long-term), two stock funds (moderate and high-risk), and a balanced fund.

02

Compute the probability

a.

The percentage of customer who owns shares in the balanced funds is provided as 7%.

The probability that the selected individual owns shares in the balanced fund is given as,

\(\frac{7}{{100}} = 0.07\)

Therefore, the probability that the selected individual owns shares in the balanced funds is 0.07.

b.

The percentage of customer who owns shares in the short bond is provided as 15%.

The percentage of customer who owns shares in the intermediate bond is provided as 10%.

The percentage of customer who owns shares in the long bond is provided as 5%.

The probability that the selected individual owns shares in a bond fund is given as,

\(\begin{aligned}\frac{{15}}{{100}} + \frac{{10}}{{100}} + \frac{5}{{100}} &= \frac{{30}}{{100}}\\ &= 0.30\end{aligned}\)

Therefore, the probability that the selected individual owns shares in a bond fund is 0.30.

c.

The percentage of customer who owns shares in the high-risk stock fund is provided as 18%.

The percentage of customer who owns shares in the moderate-risk fund is provided as 25%.

The probability that the selected individual owns shares in a stock fund is given as,

\(\begin{aligned}\frac{{18}}{{100}} + \frac{{25}}{{100}} &= \frac{{43}}{{100}}\\ &= 0.43\end{aligned}\)

The probability that the selected individual does not owns shares in a stock fund is given as,

\(1 - 0.43 = 0.57\)

Therefore, the probability that the selected individual does not owns shares in a stock fund is 0.57.

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Most popular questions from this chapter

a. A lumber company has just taken delivery on a shipment of \({\rm{10,000 2 \times 4}}\)boards. Suppose that 20% of these boards (\({\rm{2000}}\)) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let A 5 {the first board is green} and B 5 {the second board is green}. Compute \({\rm{P(A),P(B)}}\), and \({\rm{P(A}} \cap {\rm{B)}}\) (a tree diagram might help). Are \({\rm{A}}\) and \({\rm{B}}\) independent?

b. With \({\rm{A}}\) and \({\rm{B}}\) independent and \({\rm{P(A) = P(B) = }}{\rm{.2,}}\) what is \({\rm{P(A}} \cap {\rm{B)}}\)? How much difference is there between this answer and \({\rm{P(A}} \cap {\rm{B)}}\)part (a)? For purposes of calculating \({\rm{P(A}} \cap {\rm{B)}}\), can we assume that \({\rm{A}}\)and \({\rm{B}}\) of part (a) are independent to obtain essentially the correct probability?

c. Suppose the shipment consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \({\rm{P(A}} \cap {\rm{B)}}\)? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \({\rm{P(A}} \cap {\rm{B)}}\)?

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