High school conditional probability

[synkk]

Two coins are flipped and the results are recorded. Given that one coin lands on a head, find the probability of:

a) Two heads, b) a head and a tail

Searching online is giving my answers which are not using conditional probability at all, and our teacher told us we have to use the formulas etc, but i keep getting it incorrect.

Well i haven't attempted b as i can't do a but here is my working:

Let a be the coin that lands on heads
Let b be the unknown coin

so P(B | A) = P(B and A) / P(A)

P(A) = 1/2
P(B) = 1/2

P(B and A) = 1/4

P(B | A) = 1/4 / 1/2 = 1/2 which is obviously wrong.

To be honest i think P(A) is correct but i don't know how to do P(B).

 

[mathman]

Let A be event either coin lands on head.

For a. let B be the event both coins land on heads,
P(A)=3/4, P(B)=1/4, P(B|A)=(1/4)/(3/4)=1/3. Note that A ∩ B = B

For b. Let B' be the event that one coin is heads and the other is tail.
P(A)=3/4, P(B')=1/2, P(B'|A)=2/3. Here A ∩ B' = B'

 

[mathman]

Let A be event either coin lands on head.

For a. let B be the event both coins land on heads,
P(A)=3/4, P(B)=1/4, P(B|A)=(1/4)/(3/4)=1/3. Note that A ∩ B = B

For b. Let B' be the event that one coin is heads and the other is tail.
P(A)=3/4, P(B')=1/2, P(B'|A)=2/3. Here A ∩ B' = B'

Clarification: B' is NOT the complement of B in this example.

 

[synkk]

Let A be event either coin lands on head.

For a. let B be the event both coins land on heads,
P(A)=3/4, P(B)=1/4, P(B|A)=(1/4)/(3/4)=1/3. Note that A ∩ B = B

For b. Let B' be the event that one coin is heads and the other is tail.
P(A)=3/4, P(B')=1/2, P(B'|A)=2/3. Here A ∩ B' = B'

Hello,

thank you for your response i cleared it up with my teacher today though i still have a question on how to get P(AnB), why is it = to B? I thought the formula for P(AnB) = P(A) + P(B) - P(AuB), but how do you get p(AuB).

Thank you.

 

[micromass]

Hello,

thank you for your response i cleared it up with my teacher today though i still have a question on how to get P(AnB), why is it = to B? I thought the formula for P(AnB) = P(A) + P(B) - P(AuB), but how do you get p(AuB).

Thank you.

In this case, we have that $B\subseteq A$. Indeed: if both coins land on head, then one of the coins lands on head.

 

[synkk]

In this case, we have that $B\subseteq A$. Indeed: if both coins land on head, then one of the coins lands on head.

I'm sorry but what is that symbol, we haven't learned it yet.

 

[HallsofIvy]

"subset". The set B= {hh} (both heads) is a subset of A= {hh, ht, th} (at least one head).

 

[mathman]

Hello,

thank you for your response i cleared it up with my teacher today though i still have a question on how to get P(AnB), why is it = to B? I thought the formula for P(AnB) = P(A) + P(B) - P(AuB), but how do you get p(AuB).

Thank you.

Since B is a subset of A, the intersection is B and the union is A.