I have two coins: a quarter and a nickel. The quarter has a bias of 1/4 for heads, and the nickel has a bias of 2/3 for heads. I flip each coin 10 times. What is the probability that I observe exactly five heads total? Some explanation of the solution would be appreciated.
To find the probability of observing exactly five heads total when flipping both coins 10 times, we can use the concept of probability and combinatorics.
First, let's determine the number of ways we can get five heads in 10 flips. We can consider two scenarios:
1. Case 1: Getting 5 heads with the quarter and 0 heads with the nickel.
2. Case 2: Getting 4 heads with the quarter and 1 head with the nickel.
Case 1:
The probability of getting 5 heads with the quarter is given by (1/4)^5. Since the probability of getting tails is 1 - (1/4) = 3/4, the probability of getting 0 heads with the nickel is (3/4)^5. To calculate the overall probability of Case 1 occurring, we multiply these two probabilities: (1/4)^5 * (3/4)^5.
Case 2:
The probability of getting 4 heads with the quarter is given by (1/4)^4. Since the probability of getting tails is 1 - (1/4) = 3/4, the probability of getting 1 head with the nickel is (2/3)*(1/3)^4 (assuming the nickel is independent of the quarter and has a bias of 2/3 for heads). To calculate the overall probability of Case 2 occurring, we multiply these probabilities: (1/4)^4 * (2/3)*(1/3)^4.
Next, we calculate the number of ways these events can occur. For Case 1, there is only one way, as we want all 5 heads from the quarter and 0 heads from the nickel. For Case 2, there are 10 possible positions where the 1 head from the nickel could occur.
Finally, we sum up the probabilities of both cases:
Probability = (Number of ways Case 1 can occur * Probability of Case 1) + (Number of ways Case 2 can occur * Probability of Case 2)
Probability = (1 * (1/4)^5 * (3/4)^5) + (10 * (1/4)^4 * (2/3)*(1/3)^4)
Calculating this equation will give you the desired probability of observing exactly five heads total when flipping the two coins 10 times.