Calculate the probability if one of friends doesn't bring the money, then he would ask you for a beer with 90%.

Question

Let me tell you, this is not my "homework". I am preparing for the exam so I don't know the exact answer that's why I am looking for the help.

You have two friends, each having (independently) probability 0.8 of bringing money to a party. If at least one of them does not bring money, they will ask you for a beer with probability 0.9. If they both bring money, they may still ask you for a beer with probability 0.2.

1. What is the probability that they ask for a beer?
2. If they did not ask for a beer, what is the probability of both of them bringing money?
3. Generalising from 2 to n friends, how many friends do you need to have a 97% probability that at least one of them brings money? How many friends to you need to have a 97% probability that they ask you for a beer?

I am trying to calculate the point a but not sure what is the result.

So far I have tried:

x_1 = bring money
x_2 = bring money
P(x_1|x_2) = .2
P(not x_1|x_2) = .9
P(x_1|not x_2) = .9
P(not x_1| not x_2) = .9

\frac{.9*.9*.9*.2}{4} = .03645 not sure if the result is correct.

Other way I have tried:

.64 * .2 + 0.9*.36 = .452

Answer 1

• Step 1: Understand the values given to us in the problem statement and translate this into symbols which we may later manipulate.

Let $x_1$ be the event that our first friend brought money with him. Let $x_2$ be the event that our second friend brought money with him.

We are told that $x_1,x_2$ are independent from one another from the problem statement. We are further told that $Pr(x_1)=Pr(x_2)=0.8$.

Let $A$ be the event that we were asked to buy our friends some beer.

We are told that if at least one of our friends didn't bring enough money, then they will ask us to buy them a beer with probability $0.9$. Reworded, this says $Pr(A\mid x_1^c\cup x_2^c)=0.9$

Further, we were told that in the case that all of our friends brought enough money, they would still ask us to buy them a beer despite that with probability $0.2$. Reworded, this says $Pr(A\mid x_1\cap x_2)=0.2$

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• Step 2: Understand what the question is asking for

Part (a) asks us to find the probability that we are asked to buy a beer for our friends, i.e. to find $Pr(A)$.

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• Step 3: Understand what information we have and how it relates to the information we want

Remember that $Pr(A) = Pr(A\cap B)+Pr(A\cap B^c) = Pr(B)Pr(A\mid B) + Pr(B^c)Pr(A\mid B^c)$, a special case of the "Law of Total Probability." Now, using $x_1\cap x_2$ in place of $B$ and $x_1^c\cup x_2^c$ in place of $B^c$, we can continue.

$Pr(A)=0.8\cdot 0.8\cdot 0.2 + (1- 0.8\cdot 0.8)\cdot 0.9$

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The later parts follow similarly. Note that you will also want to use Bayes Theorem $Pr(A\mid B) = \dfrac{Pr(B\mid A)Pr(A)}{Pr(B)}$ and you will want to generalize the formulas to $n$ friends instead. All that generalizing to $n$ friends will change is the probability that at least one forgot his money versus all having brought money.