Suppose that we have two coins C1 and C2. The probability of C1 having head is 0.6, and the probability of C2 having head is 0.4. In each test, we toss both coins, and read the faces of C1 and C2 (note that we read C2 after reading C1). For example, if the toss resulted in C1 head up and C2 tail up, we will record the result as HT. Suppose we perform the test 4 times. What is the probability for us to observe the following result? HT, HT, TT, TT?
My answer is actually crossing 1 and this couldn't be right I guess :
- First Test with H on C1 and T on C2, the Probability is 0.6 * 0.6 = 0.36
- Second Test with H on C1 and T on C2, the probability is 0.6 * 0.6 = 0.36
- Third test with T on C1 and T on C2, the probability is 0.4 * 0.6 = 0.24
- Fourth test with T on C1 and T on C2, the probability is 0.4 * 0.6 = 0.24
So the total probability comes out to be 1.2 which doesn't make any sense. I initially thought of solving this using the Bayes theorem but not sure how to apply conditional probability over here.