# Math Help - Conditional probability

1. ## Conditional probability

Okay, so I'm stuck on this question:

You decide to play a game. The rules for the game is that there are 8 question that you need to answer. For each question, you are given 4 different options for answer.
Before you start a game, you have to place a starting bid. When all 8 questions have been answered, the following prizes are given:

If all 8 questions have been answered correctly, you get 1000 times your starting bid back.
If 7 questions have been answered correctly, you get 100 times your starting bid back.
If 6 questions have been answered correctly, you get 10 times your starting bid back.
5 correct answers and below, does not get you any prizes.

You use a starting bid of $100. How many questions would you have to know the answer to, before you in average get a profit from this game? So basically, I believe that you need to consider the fact that it's a conditional probability, since the chance of getting a prize goes up for every question that you know the answer to. But I just can't figure this out. 2. Originally Posted by No Logic Sense Okay, so I'm stuck on this question: You decide to play a game. The rules for the game is that there are 8 question that you need to answer. For each question, you are given 4 different options for answer. Before you start a game, you have to place a starting bid. When all 8 questions have been answered, the following prizes are given: If all 8 questions have been answered correctly, you get 1000 times your starting bid back. If 7 questions have been answered correctly, you get 100 times your starting bid back. If 6 questions have been answered correctly, you get 10 times your starting bid back. 5 correct answers and below, does not get you any prizes. You use a starting bid of$100.
How many questions would you have to know the answer to, before you in average get a profit from this game?

So basically, I believe that you need to consider the fact that it's a conditional probability, since the chance of getting a prize goes up for every question that you know the answer to. But I just can't figure this out.
Assume first that you guess the answers.

Let X be the random variable number of correct questions.

X ~ Binomial(n = 8, p = 1/4).

Calculate Pr(X = 8), Pr(X = 7), Pr(X = 6) and Pr(X < 6).

Now calculate the expected profit. What does this tell you about the minimum number of questions you need to know the answer to so that the expected profit is greater than zero?

3. Well, I calculated the probability of guessing 8 correct, 7 correct and 6 correct, by using binomial probability:

$p=1/4$
$1-p=3/4$
$n=8$

$P(8)=b(8;8,1/4)={\left ( \frac{1}{4} \right )}^8=\frac{1}{65536}$

$P(7)=b(7;8,1/4)=\binom{8}{7}\cdot{\left ( \frac{1}{4} \right )}^7\cdot{\left ( \frac{3}{4} \right )}^{8-7}=\frac{3}{8192}$

$P(6)=b(6;8,1/4)=\binom{8}{6}\cdot{\left ( \frac{1}{4} \right )}^6\cdot{\left ( \frac{3}{4} \right )}^{8-6}=\frac{63}{16384}$

I'm not sure I understand the X < 6 part. I mean, under 6 correct answers does not give you any prize, so it doesn't have anything to do with the profit.

4. Originally Posted by No Logic Sense
Well, I calculated the probability of guessing 8 correct, 7 correct and 6 correct, by using binomial probability:

$p=1/4$
$1-p=3/4$
$n=8$

$P(8)=b(8;8,1/4)={\left ( \frac{1}{4} \right )}^8=\frac{1}{65536}$

$P(7)=b(7;8,1/4)=\binom{8}{7}\cdot{\left ( \frac{1}{4} \right )}^7\cdot{\left ( \frac{3}{4} \right )}^{8-7}=\frac{3}{8192}$

$P(6)=b(6;8,1/4)=\binom{8}{6}\cdot{\left ( \frac{1}{4} \right )}^6\cdot{\left ( \frac{3}{4} \right )}^{8-6}=\frac{63}{16384}$

I'm not sure I understand the X < 6 part. I mean, under 6 correct answers does not give you any prize, so it doesn't have anything to do with the profit.
Under 6 correct answers means you lose your starting bid, doesn't it? So the profit is -\$100.

5. I see. Yes, I agree now.

So, to calculate this, wouldn't that be P(5)+P(4)+P(3)+P(2)+P(1)?

6. Originally Posted by No Logic Sense
I see. Yes, I agree now.

So, to calculate this, wouldn't that be P(5)+P(4)+P(3)+P(2)+P(1)?
+ P(0).

7. Originally Posted by mr fantastic
+ P(0).
Yes of course.
So:

$P(5)=\frac{189}{8192}
P(4)=\frac{2835}{32768}
P(3)=\frac{1701}{8192}
P(2)=\frac{5103}{16384}
P(1)=\frac{2187}{8192}
P(0)=\frac{6561}{65536}$

And the sum will be:

$P(X<6)=P(5)+P(4)+P(3)+P(2)+P(1)+P(0)=\frac{65259}{ 65536}$

Which almost equals 1.

Back to the expected profit, it is the average profit, right?

8. Originally Posted by No Logic Sense
Yes of course.
So:

$P(5)=\frac{189}{8192}
P(4)=\frac{2835}{32768}
P(3)=\frac{1701}{8192}
P(2)=\frac{5103}{16384}
P(1)=\frac{2187}{8192}
P(0)=\frac{6561}{65536}$

And the sum will be:

$P(X<6)=P(5)+P(4)+P(3)+P(2)+P(1)+P(0)=\frac{65259}{ 65536}$

Which almost equals 1.

Back to the expected profit, it is the average profit, right?
Yes.

Now, having done all this, what you'll probably need to do now is work things out assuming that he knows the answers to say, 2 questions. So calculate Pr(X = 6) (if X = 6 then he gets the correct answer to 8 questions since he already knows 2 of them), Pr(X = 7) (if X = 5 then he gets the correct answer to 7 questions snce he already knows 2 of them) , Pr(X = 4) and Pr(X < 4).

You might need to play around with how many answers are known until you get an expected profit that is greater than zero.