Solving a Coin Tossing Problem with Probability

  • Context: Undergrad 
  • Thread starter Thread starter viviane363
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Discussion Overview

The discussion revolves around calculating the probability of obtaining heads when randomly selecting and tossing one of several coins with different characteristics. The coins include three double-headed coins, four double-tailed coins, and two fair coins. Participants explore various methods to approach the problem, including conditional probabilities and direct counting of outcomes.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • Some participants propose defining events A, B, and C for the different types of coins and calculating probabilities based on these definitions.
  • One participant suggests using the law of total probability to express P(H) as a sum of conditional probabilities.
  • Another participant calculates P(H) using specific probabilities for each coin type but later questions the initial probability assignments for A, B, and C.
  • Some participants emphasize the importance of ensuring that the probabilities for selecting each coin type sum to 1.
  • One participant introduces the idea of using a probability tree to visualize the problem and suggests it could help clarify the solution.
  • Another participant calculates the probability of heads by considering the total number of heads and sides among the coins, arriving at a probability of 4/9.
  • There is a discussion about whether the problem should be approached using conditional probabilities or direct counting methods, with differing opinions on the best approach.

Areas of Agreement / Disagreement

Participants generally agree that the probabilities must sum to 1, but there is disagreement on the correct assignments of these probabilities and the best method to solve the problem. Multiple competing views remain on how to approach the calculation of the probability of heads.

Contextual Notes

Some participants note that the initial probability assignments for the events A, B, and C may be incorrect, leading to confusion in the calculations. There is also mention of the need to properly weight the populations of coins when combining probabilities.

viviane363
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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.
 
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viviane363 said:
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?
 
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?
 
Your probabilities for the events A, B, and C can't be right. Hint: they must sum to 1.
 
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?
 
viviane363 said:
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
 
viviane363 said:
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.
 
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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 said:
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?
 
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  • #10
SW VandeCarr said:
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?
 
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  • #11
bpet said:
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.
 
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