Conditional Probability on type of coin

In summary, when one of the coins is randomly selected and flipped, the conditional probability that it was the coin i is 1/10.
  • #1
CAF123
Gold Member
2,948
88

Homework Statement


Suppose we have 10 coins such that if the ith coin is flipped, heads will appear with probability i/10, i = 1,2...10. When one of the coins is randomly selected and flipped, what is the conditional probability that it was the coin?


Homework Equations


Bayes's Formula


The Attempt at a Solution


First of all, I can't make any sense out of how choosing a different coin will give a different probability of showing a head?
In solving the problem, I used Bayes's relation; [tex] P(\text{5th coin} | \text{heads}) = \frac{P(\text{heads | 5th coin})P(\text{5th coin})}{P(\text{heads| 5th coin})P(\text{5th coin}) + P(\text{heads | not 5th coin})P(\text{not 5th coin})} [/tex]

where [tex] P(\text{heads |5th coin}) = \frac{5}{10}, P(\text{5th coin}) = \frac{1}{10}, P(\text{heads | not 5th coin}) = \frac{P(\text{heads and not 5th coin})}{P(\text{not 5th coin})} = \frac{\frac{9}{10}\frac{1}{2}}{\frac{9}{10}} = \frac{1}{2}, P(\text{not 5th coin}) = 1-\frac{1}{10}. [/tex] Putting this together gives the wrong answer. Any ideas?
 
Physics news on Phys.org
  • #2
CAF123 said:

Homework Statement


Suppose we have 10 coins such that if the ith coin is flipped, heads will appear with probability i/10, i = 1,2...10. When one of the coins is randomly selected and flipped, what is the conditional probability that it was the coin?


Homework Equations


Bayes's Formula


The Attempt at a Solution


First of all, I can't make any sense out of how choosing a different coin will give a different probability of showing a head?
In solving the problem, I used Bayes's relation; [tex] P(\text{5th coin} | \text{heads}) = \frac{P(\text{heads | 5th coin})P(\text{5th coin})}{P(\text{heads| 5th coin})P(\text{5th coin}) + P(\text{heads | not 5th coin})P(\text{not 5th coin})} [/tex]

where [tex] P(\text{heads |5th coin}) = \frac{5}{10}, P(\text{5th coin}) = \frac{1}{10}, P(\text{heads | not 5th coin}) = \frac{P(\text{heads and not 5th coin})}{P(\text{not 5th coin})} = \frac{\frac{9}{10}\frac{1}{2}}{\frac{9}{10}} = \frac{1}{2}, P(\text{not 5th coin}) = 1-\frac{1}{10}. [/tex] Putting this together gives the wrong answer. Any ideas?

There is something missing from your problem statement: you say "When one of the coins is randomly selected and flipped, what is the conditional probability that it was the coin?" Did you mean that one of the coins was flipped and came up heads? Did you mean 'what is the conditional probability it was coin i?" I will assume the answer is YES to both of these questions.

P{coin i|H} = P{coin i & H}/P{H}. What is the numerator equal to? How do you find the denominator?

RGV
 
  • #3
Oh sorry, how careless of me. It should read 'When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the 5th coin'
 
  • #4
CAF123 said:
Oh sorry, how careless of me. It should read 'When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the 5th coin'

OK, so my previous post is correct if we put i = 5. Now can you answer the questions I asked there? Take it one step at a time. Apply Bayes' rules, etc.

RGV
 
  • #5
Ray Vickson said:
OK, so my previous post is correct if we put i = 5. Now can you answer the questions I asked there? Take it one step at a time. Apply Bayes' rules, etc.

RGV

I got that P(coin 5|head) = P(head|coin5)P(coin5)/ƩP(head|coin i)P(coin i). (Where the sum is from i=1 to i=5). Is this correct so far?
 
  • #6
CAF123 said:
I got that P(coin 5|head) = P(head|coin5)P(coin5)/ƩP(head|coin i)P(coin i). (Where the sum is from i=1 to i=5). Is this correct so far?

We have 10 coins, not 5.

RGV
 
  • #7
I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said [itex] \sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) + (6/10)(1/10) + (7/10)(1/10) + (8/10)(1/10) + (9/10)(1/10) + (1/10) [/itex] I now get the right result - thanks for your help!

EDIT : let i go form 1 to 10.
 
Last edited:
  • #8
CAF123 said:
I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said [itex] \sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) [/itex] This is still incorrect.

What's so special about the number '5'? Suppose, instead, I asked you for the conditional probability of coin 2, or coin 7 or coin 10. How would you express the conditional probabilities in those cases? Remember, I first asked you about the conditional probability of coin i, where I did not specify i.

RGV
 
  • #9
CAF123 said:
I think P(head/coin 5) = 5/10 and P(coin i ) = 1/10 since all equally likely to be picked. I then said [itex] \sum_{i=1}^{5} P(head|coin\, i)P(coin\, i) = (1/10)(1/10) + (2/10)(1/10) + (3/10)(1/10) + (4/10)(1/10) + (5/10)(1/10) + (6/10)(1/10) + (7/10)(1/10) + (8/10)(1/10) + (9/10)(1/10) + (1/10) [/itex] I now get the right result - thanks for your help!

EDIT : let i go form 1 to 10.

OK now, but of course you need to say ##\sum_{i=1}^{10},## which is actually what you calculated.

RGV
 

1. What is conditional probability?

Conditional probability is the likelihood of an event occurring based on the occurrence of another event.

2. How is conditional probability applied to different types of coins?

Conditional probability can be used to determine the likelihood of obtaining a specific outcome when flipping different types of coins, such as fair coins, biased coins, or weighted coins.

3. What factors affect the conditional probability of a certain coin type?

The conditional probability of a certain coin type can be influenced by factors such as the weight distribution of the coin, the surface texture of the coin, and the flipping technique used.

4. How is conditional probability calculated for coin flipping experiments?

Conditional probability for coin flipping experiments can be calculated by dividing the number of desired outcomes by the total number of possible outcomes.

5. Can conditional probability be used to predict the outcome of a coin flip?

Yes, conditional probability can be used to make predictions about the likelihood of obtaining a certain outcome when flipping a coin, based on the characteristics of the coin and the method of flipping.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
4
Views
3K
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
798
  • Precalculus Mathematics Homework Help
Replies
16
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Set Theory, Logic, Probability, Statistics
2
Replies
57
Views
1K
Back
Top