Question

1. claire is considering investing in a new business. in the first year, there is a probability of 0.2 that the new business will lose $10,000, a probability of 0.4 that the new business will break even ($0 loss or gain), a probability of 0.3 that the new business will make $5,000 in profits, and a probability of 0.1 that the new business will make $8,000 in profits.

a. claire should invest in the company if she makes a profit. should she invest? explain using expected values.
b. if claires initial investment is $1,200 and the expected value for the new business stays constant, how many years will it take for her to earn back her initial investment?

Explanation:

Step1: Calculate the expected value for part a

The formula for expected value \(E(X)=\sum_{i = 1}^{n}x_ip_i\).
Let \(x_1=- 10000\), \(p_1 = 0.2\); \(x_2 = 0\), \(p_2=0.4\); \(x_3 = 5000\), \(p_3=0.3\); \(x_4 = 8000\), \(p_4=0.1\).

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Step2: Analyze the result for part a

Since the expected value \(E(X) = 300>0\), on average, Claire can make a profit.

Step3: Solve part b

Let \(n\) be the number of years. We know that the initial investment is \(I = 1200\) and the annual expected profit is \(E(X)=300\).
We use the formula \(I=n\times E(X)\), so \(n=\frac{I}{E(X)}\).
Substitute \(I = 1200\) and \(E(X)=300\) into the formula: \(n=\frac{1200}{300}=4\)

Answer:

a. Claire should invest. The expected value of the investment is \(\$300\) (calculated as \((-10000)\times0.2+0\times0.4 + 5000\times0.3+8000\times0.1 = 300\)), which means on average she will make a profit.
b. It will take \(4\) years. (Using the formula \(n=\frac{I}{E(X)}\), with \(I = 1200\) and \(E(X)=300\), \(n=\frac{1200}{300}=4\))