1. ## Basic Probability help

Ok, here's the problem.

The probability of a child inheriting a certain disease is 1/2. In a two child family, what is the probability that the younger child will inherit this disease given that the older child has already inherited it?

Here's what I've come up with, but I don't feel that it's right.

Events:
A=Older child is affected B=Younger child is affected
Looking for:
P(B|A)
Know:
P(A) = 1/2 P(B) = 1/2
P(B│A)= (P(B∩A))/(P(A))= (1/4)/(1/2)=2/4=1/2

I guess this makes sense because the probability of each child getting the disease is independent of the other one having the disease but a part of me feels that the answers should be 1/4 because according to the problem, the older child already has the disease meaning that P(A) = 1

2. Originally Posted by downthesun01
Ok, here's the problem.

The probability of a child inheriting a certain disease is 1/2. In a two child family, what is the probability that the younger child will inherit this disease given that the older child has already inherited it?

Here's what I've come up with, but I don't feel that it's right.

Events:
A=Older child is affected B=Younger child is affected
Looking for:
P(B|A)
Know:
P(A) = 1/2 P(B) = 1/2
P(B│A)= (P(B∩A))/(P(A))= (1/4)/(1/2)=2/4=1/2

I guess this makes sense because the probability of each child getting the disease is independent of the other one having the disease but a part of me feels that the answers should be 1/4 because according to the problem, the older child already has the disease meaning that P(A) = 1
Let me ask you a question:

The probability of a coin landing heads is 1/2. In two coin tosses, what is the probability that the first toss will be a head given that the second toss is a head?

3. Honestly, I'm not even sure at this point. On the one hand I would say it's 1/2 because the second flip has no influence on the first flip. But on the other hand, 1/4 seems more likely because E = {HH, HT, TH, TT} and if the second flip has already been determined to be heads, then the only way the first flip can be heads is if the total outcome was {HH}

As you can see, I'm really, really confusing myself with this. 1/4 makes the most sense to me, but I can't see to prove it mathematically.

Edit: Wait, I think I just get it. After looking at what I just typed, the answer does seem to be 1/2 because once it's established that the second flip is heads events {HT} and {TT} get removed since the second flip is tails in both of those, leaving only events {HH} and{TH} as possible outcomes. With these as the only two possible outcomes, it's obvious that the first flip has a 1/2 chance of hitting heads. Is this correct?

4. Originally Posted by downthesun01
Honestly, I'm not even sure at this point. On the one hand I would say it's 1/2 because the second flip has no influence on the first flip. But on the other hand, 1/4 seems more likely because E = {HH, HT, TH, TT} and if the second flip has already been determined to be heads, then the only way the first flip can be heads is if the total outcome was {HH}

As you can see, I'm really, really confusing myself with this. 1/4 makes the most sense to me, but I can't see to prove it mathematically.

Edit: Wait, I think I just get it. After looking at what I just typed, the answer does seem to be 1/2 because once it's established that the second flip is heads events {HT} and {TT} get removed since the second flip is tails in both of those, leaving only events {HH} and{TH} as possible outcomes. With these as the only two possible outcomes, it's obvious that the first flip has a 1/2 chance of hitting heads. Is this correct?
If the second toss is a head then the reduced sample space is HH and TH and each of these events is equally likely. Therefore probability of a head in the first toss given a head in the second toss is 1/2.

5. Originally Posted by mr fantastic
If the second toss is a head then the reduced sample space is HH and TH and each of these events is equally likely. Therefore probability of a head in the first toss given a head in the second toss is 1/2.
Cool thanks. It didn't come to me until I wrote out the sample space and realized that by the second flip equaling heads all events in the sample space in which the second flip equaled tails are eliminated leaving only two outcomes.

Can someone comment on whether this is a mathematically valid way to answer my original question?

Events:
A=Older child is affected B=Younger child is affected
Looking for:
P(B|A)
Know:
P(A) = 1/2 P(B) = 1/2
P(B│A)= (P(B∩A))/(P(A))= (1/4)/(1/2)=2/4=1/2

6. Originally Posted by downthesun01
Cool thanks. It didn't come to me until I wrote out the sample space and realized that by the second flip equaling heads all events in the sample space in which the second flip equaled tails are eliminated leaving only two outcomes.

Can someone comment on whether this is a mathematically valid way to answer my original question?

Events:
A=Older child is affected B=Younger child is affected
Looking for:
P(B|A)
Know:
P(A) = 1/2 P(B) = 1/2
P(B│A)= (P(B∩A))/(P(A))= (1/4)/(1/2)=2/4=1/2
I still think you're going about this the hard way. Since events A and B are independent it's sufficient to say Pr(B|A) = Pr(B) = 1/2.

7. I gotcha, thanks. One of the follow up questions asked whether A and B are independent and I put since:

P(B│A)= (P(B∩A))/(P(A)) = P(B) = 1/2

They are.
Then I started second-guessing myself into thinking that maybe the probability was somehow 1/4. Anyway, I may have a couple more questions but I've got class today. Hopefully they'll get answered in class and I won't have to spam your forums with such trivial stuff.

Thanks again. You were a big help.

8. Originally Posted by downthesun01
I gotcha, thanks. One of the follow up questions asked whether A and B are independent and I put since:

P(B│A)= (P(B∩A))/(P(A)) = P(B) = 1/2

They are.
Then I started second-guessing myself into thinking that maybe the probability was somehow 1/4. Anyway, I may have a couple more questions but I've got class today. Hopefully they'll get answered in class and I won't have to spam your forums with such trivial stuff.

Thanks again. You were a big help.
You're welcome.

And please don't consider your posts as trivial or spam. The discussion in this thread has helped you and it will help others who read it. Hardly trivial and hardly the stuff of spam ....