# Please check my Bayes Theorem and Independent Events Probability Exercises

I'm doing an exercise that I think is conditional probability and hence must be used Bayes Theorem. I just need to check with the help of you guys if my analysis is correct. I will start by stating the exercise:

An investor is thinking of buying a very large number of shares of a company. The stock price on the stock exchange during the previous six months is of great interest to the investor. Based on this information, she observes that the price is related to the gross national product (GNP). If the GNP increases, the probability that the share price increases is 0.7. If the GNP is the same, the probability that the shares increase their value is 0.2, while if the GNP decreases then the probability that the shares increase their value is only 0.1. The probability that the GNP increases, remains the same or decreases are respectively 0.5, 0.3 and 0.2.

I did a tree diagram to better visualize this situation. See the attached picture.

1. What is the probability that the shares increase their value?

$$P(shares.increase)=0.35 + 0.06 + 0.02 = 0.43$$

1. If the shares increased their value, what is the probability that the GNP has increased?

$$P(GNP.increased | shares.increased)$$

$$=\frac{P(GNP.increased)*P(shares.increased | GNP. increased)}{P(GNP.inc)*P(shares.inc | GNP.inc)+P(GNP.same)*P(shares.inc | GNP.same)+P(GNP.dec)*P(shares.inc | GNP.dec)}$$

$$=\frac{0.35}{0.35+0.06+0.02}$$

$$=\frac{0.35}{0.43}$$

$$=\frac{0.35}{0.43}$$

$$≈0.81$$

I have another exercise of probability that I would be appreciated if you can check it out. The exercise says:

Let A, B and C independents events with $$P(A)=P(B)=P(C)=\frac{1}{3}$$. Calculate the probability that:

a) At least one of the events happen:

So if $$P(A)=P(B)=P(C)=\frac{1}{3}$$, then $$P(A^c)=P(B^c)=P(C^c)=\frac{2}{3}$$

$$P(at.least.one.of.the.events.happen)$$ $$= 1 - P (none.of.them.happen)$$ $$= 1 - (\frac{2}{3})^3$$ $$= 1 - \frac{8}{27}$$ $$= \frac{19}{27}$$

b) At least two of them happen:

$$P(at.least.two.of.them.happen)$$

$$=1-P(none.of.them.happen)-P(exactly.one.of.them.happen)$$

$$=1-\frac{8}{27}-(\frac{1}{3}\cdot\frac{2}{3}\cdot\frac{2}{3}+\frac{2}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}+\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3})$$

$$=1-\frac{8}{27}-\frac{4}{9}$$

$$=\frac{7}{27}$$

c) Exactly two of them happen:

$$P(Exactly.two.of.them.happen)$$

$$=\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}+\frac{1}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}+\frac{2}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}$$

$$=\frac{2}{9}$$

Thanks for taking the time to check. If something is wrong, I will be thankful to be corrected. If I'm right, and there is another easy way to do any of these, I'll be happy to see another way. Thanks thanks.

• Hi, I know this question is about a year old, but you still haven't accepted my answer, I'd appreciate it if you did! – uriyaba Dec 7 '20 at 13:04

## 1 Answer

All of your answers are correct, and you've answered them in a very efficient way.

I don't see an easier way of making these calculations, other than shortening the answer like so:

• $$(\frac13\cdot\frac23\cdot\frac23+\frac23\cdot\frac13\cdot\frac23+\frac23\cdot\frac23\cdot\frac13) = 3\cdot(\frac13\cdot\frac23\cdot\frac23)$$
• $$(\frac13\cdot\frac13\cdot\frac23+\frac13\cdot\frac23\cdot\frac13+\frac23\cdot\frac13\cdot\frac13) = 3\cdot(\frac13\cdot\frac13\cdot\frac23)$$
• Thanks so much. – gi2302 Dec 3 '19 at 22:52
• You're welcome! I'd appreciate it if you ticked my answer :) – uriyaba Dec 13 '19 at 12:21