# Thread: Bayes Theorem Problem Sets

1. ## Bayes Theorem Problem Sets

Hi there! Hope you all could help me out with the following problems (4 in total):

1. It is known that a certain disease is caused by a gene mutation but is recessive in nature. This means that if B is a normal gene, and b is the disease gene, only people with genotype bb will get the disease. It is also known that in a certain population 2.25% of the population has the disease. Assume that the population is in a Hardy-Weinberg equilibrium. If 2 carriers (Bb gene combi) have children, their children have a 1 in 4 chance of contracting the disease.

Given that a man and his wife do not have the disease, what is the probability that at least one of his first two children will have the disease?

I chose to do it this way:

(.25)(.25^2 + (.25x.75x2))

Is this right? My reasoning was that there are 3 types of couples: BB BB, Bb Bb, and BB Bb. So there is a 1 in 4 chance that we have the Bb Bb couple (the others will not have children who have the disease...at most only carriers). Then with regards to the carrier couple, there are 2 scenarios...1 where both children have the disease and the other where only 1 has it.

2. Suppose that we want to determine what percentage of the students at a large university cheated in the last year examination. We construct 20 cards, with "I cheated in last year's exam" written on 4 of them and "I did not cheat in last year's exam" written on 16 of them. We then let each student chosen in the sample interview select one of the 20 cards at random and respond "correct" or "incorrect" without divulging the statement. If 1540 out of 200 students answered "correct" in the sample interview, what is the probability that a student chosen at random in the university would have cheated in last year's exam?

I used the randomised response technique formula for this. So I got the answer 0.95. I am just seeking confirmation for this one...

3. A and B play a series of games. Each game is independently won by A with probability of 0.4 and by B with a probability of 0.6.

(a) They stop when the total number of wins of one of the players is 2 greater than that of the other. The player with the larger number of total wins is the match winner.

(i) what is the probability that a total of 4 games were played?

I drew a tree diagram for this and got 2(.4^3 x .6) + 2(.6^3 x .4) = .2496

(ii) what is the probability that A is the match winner?

this one I am not sure how to approach

(b) If A and B stop when the total number of wins of one of the players is 4 greater than that of the other, and the player with the larger number of wins is the match winner, then what is the probability that A is the winner?

I think if I know how to approach (aii) I'll know how to approach this one.

2. Originally Posted by deathstarx
Hi there! Hope you all could help me out with the following problems (4 in total):

1. It is known that a certain disease is caused by a gene mutation but is recessive in nature. This means that if B is a normal gene, and b is the disease gene, only people with genotype bb will get the disease. It is also known that in a certain population 2.25% of the population has the disease. Assume that the population is in a Hardy-Weinberg equilibrium. If 2 carriers (Bb gene combi) have children, their children have a 1 in 4 chance of contracting the disease.

Given that a man and his wife do not have the disease, what is the probability that at least one of his first two children will have the disease?

I chose to do it this way:

(.25)(.25^2 + (.25x.75x2))

Is this right? My reasoning was that there are 3 types of couples: BB BB, Bb Bb, and BB Bb. So there is a 1 in 4 chance that we have the Bb Bb couple (the others will not have children who have the disease...at most only carriers). Then with regards to the carrier couple, there are 2 scenarios...1 where both children have the disease and the other where only 1 has it.
You are told that the frequency $p(bb)=0.0225$, from this you should deduce that in HW equilibrium $p(BB)=0.7225$ and $p(bB)=0.255$.

Given the parents do not have the disease we know that for a parent we we have $P(bb)=0$, $P(BB)=p(BB)/(p(BB)+p(bB))$, $P(bB)=p(bB)/(p(BB)+p(bB))$ and we have the following cases:

parent 1 BB and parent 2 BB with probability $P(BB)^2$

parent 1 BB and parent 1 bB with probability $P(BB)P(bB)$

parent 1 bB and parent 2 BB with probability $P(bB)P(BB)$

parent 1 bB and parent 2 bB with probability $P(bB)^2$

We know that the probability of any children having the disease in the first three of these cases is 0, which leaves the third case
(when the number of children who have the disease is $\sim B(2,0.25)$).

CB

3. Thanks Captain Black, but do I need to weight it by the probability of having that type of couple?

4. Originally Posted by deathstarx
Thanks Captain Black, but do I need to weight it by the probability of having that type of couple?
Yes, that's why I gave the required calculation.

CB

5. Oh. I see I see. Sorry for being a little slow, but is that notation for binomial distributions?

6. Originally Posted by deathstarx
Oh. I see I see. Sorry for being a little slow, but is that notation for binomial distributions?
There are other notations but in this case B(N,p) denotes the distribution of the number of successes in N trials where the probability of success on a single trial is p.

CB

7. Oki thanks.