Solving a Coin Tossing Problem with Probability

[viviane363]

Hi, I have this problem here that I am trying to solve:
A box contains three coins with a head on each side, four coins with a tail on each side, and two fair coins. If one of these nine coins is selected at random and tossed once, what is the probability that a head will be obtained?
Here is what I think:
1) A: the event that one of the three coins with a head on each side is chosen.
B: the event that one of the four coins with a tail on each side is chosen.
C: the event that one of two fair coins is chosen.

P(A)=1/8, P(B)=1/16, P(C)=1/4
P(a head will be obtained)= ? I don't know where to start from there now.

 

[awkward]

Hi, I have this problem here that I am trying to solve:
A box contains three coins with a head on each side, four coins with a tail on each side, and two fair coins. If one of these nine coins is selected at random and tossed once, what is the probability that a head will be obtained?
Here is what I think:
1) A: the event that one of the three coins with a head on each side is chosen.
B: the event that one of the four coins with a tail on each side is chosen.
C: the event that one of two fair coins is chosen.

P(A)=1/8, P(B)=1/16, P(C)=1/4
P(a head will be obtained)= ? I don't know where to start from there now.

Let's say H is the event of getting a head.

Then P(H) = P(H|A) P(A) + P(H|B) P(B) + P(H|C) P(C).

Can you take it from there?

 

[viviane363]

Here is what I am thinking:
P(H) = P(H|A) P(A) + P(H|B) P(B) + P(H|C) P(C)
=(1).(1/8)+(0).(1/16)+(1/2).(1/4)
=(1/8)+(1/8)
=2/8
Is it?

 

[Moo Of Doom]

Your probabilities for the events A, B, and C can't be right. Hint: they must sum to 1.

 

[viviane363]

I am a bit lost, and I am not too sure what I missed. For sure the sum of the probabilities has to be 1, but I can't figure out what I did wrong. Can you give me another hint?

 

[jambaugh]

I am a bit lost, and I am not too sure what I missed. For sure the sum of the probabilities has to be 1, but I can't figure out what I did wrong. Can you give me another hint?

How many total coins?
How many double headed coins?

P(double head) = number/total

 

[SW VandeCarr]

Hi, I have this problem here that I am trying to solve:
A box contains three coins with a head on each side, four coins with a tail on each side, and two fair coins. If one of these nine coins is selected at random and tossed once, what is the probability that a head will be obtained?
Here is what I think:
1) A: the event that one of the three coins with a head on each side is chosen.
B: the event that one of the four coins with a tail on each side is chosen.
C: the event that one of two fair coins is chosen.

P(A)=1/8, P(B)=1/16, P(C)=1/4
P(a head will be obtained)= ? I don't know where to start from there now.

For the set of double-faced coins the probability of heads is 3/7. For the set of two fair coins the probability of getting a head is 1/2 in one toss. If you combine the two sets the probability of getting a head is 7/9(3/7)+2/9(1/2)= 4/9. It doesn't matter whether any coin is tossed or or not. Since you are selecting just one coin from the combined set, this is not a conditional or joint probability.

 

[bpet]

For these sort of problems it's well worth drawing the probability tree. There are plenty of good examples on the web if it's not covered in the textbook you're working from, and it will be a great help for understanding even more complicated problems.

 

[SW VandeCarr]

For these sort of problems it's well worth drawing the probability tree. There are plenty of good examples on the web if it's not covered in the textbook you're working from, and it will be a great help for understanding even more complicated problems.

bpet.

I edited my post. I believe the answer is correct. It's not a conditional or joint probability. Any comment?

 

[bpet]

bpet.

I edited my post. I believe the answer is correct. It's not a conditional probability. Any comment?

Not sure how the 3/7 was obtained. For this problem since the chance of heads or tails depends on which coin type is chosen, it's natural to describe it in terms of conditional probabilities and a probability tree is a convenient way to visualize it. Does that get the same answer?

Edit: on second thoughts, to solve it without conditional probabilities - since there are 18 equally likely coin sides to choose from, how many of these are heads?

 

[SW VandeCarr]

Not sure how the 3/7 was obtained. For this problem since the chance of heads or tails depends on which coin type is chosen, it's natural to describe it in terms of conditional probabilities and a probability tree is a convenient way to visualize it. Does that get the same answer?

Edit: on second thoughts, to solve it without conditional probabilities - since there are 18 equally likely coin sides to choose from, how many of these are heads?

There are 7 double sided coins, three double heads and four double tails. If a double sided head is chosen, the toss must be heads. If a double sided tail is chosen, the toss cannot be heads. Hence for the double sided population p(heads)=3/7. For the fair coin population p(heads)=1/2. My initial mistake was not to properly weight the two populations when I combined them. The weights are 7/9 and 2/9 (post 7).

Your approach would work too. There are 8 heads out of a total population of 18 sides. This also gives 4/9. I admit, this is simpler. I just didn't think of it that way.