Solving a Coin Tossing Problem with Probability

  • Thread starter viviane363
  • Start date
In summary: Then the probability of heads is 8/18=4/9.In summary, the probability of getting a head when randomly selecting and tossing one of the nine coins in the box is 4/9. This can be calculated by summing the conditional probabilities for each type of coin, with a weight of 7/9 for the double-sided coins and 2/9 for the fair coins. Alternatively, the probability can be obtained by counting the total number of heads (8) out of the total number of coin sides (18), giving a probability of 4/9.
  • #1
viviane363
17
0
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.
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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?
 
  • #4
Your probabilities for the events A, B, and C can't be right. Hint: they must sum to 1.
 
  • #5
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?
 
  • #6
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
 
  • #7
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.
 
Last edited:
  • #8
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.
 
  • #9
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?
 
Last edited:
  • #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?
 
Last edited:
  • #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.
 
Last edited:

Related to Solving a Coin Tossing Problem with Probability

What is a coin tossing problem?

A coin tossing problem is a mathematical question that involves determining the probability of a certain outcome when flipping a coin. This can include questions about the likelihood of getting a certain number of heads or tails in a certain number of flips.

What is probability?

Probability is a measure of the likelihood of a certain event occurring. In the context of a coin tossing problem, it is the chance of getting a certain outcome, such as heads or tails, when flipping a coin.

How do you solve a coin tossing problem with probability?

To solve a coin tossing problem with probability, you need to first define the event you are interested in and then calculate the probability using the formula P(event) = (# of favorable outcomes) / (total # of possible outcomes). This formula can be applied to both simple and more complex coin tossing problems.

What is a fair coin?

A fair coin is one that has an equal chance of landing on heads or tails when flipped. This means that the probability of getting heads is 0.5, and the probability of getting tails is also 0.5. In a fair coin, the outcomes are independent of each other, meaning the previous flip does not affect the next one.

Why are coin tossing problems important?

Coin tossing problems are important because they allow us to understand and calculate the likelihood of certain outcomes in a random event. This has practical applications in fields such as statistics, economics, and gambling. Additionally, understanding probability can help us make informed decisions and predictions in various situations.

Similar threads

  • Set Theory, Logic, Probability, Statistics
2
Replies
57
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Set Theory, Logic, Probability, Statistics
2
Replies
41
Views
4K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
20
Views
4K
  • Set Theory, Logic, Probability, Statistics
2
Replies
45
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
11
Views
1K
  • Set Theory, Logic, Probability, Statistics
4
Replies
126
Views
7K
  • Set Theory, Logic, Probability, Statistics
Replies
21
Views
3K
Back
Top