4. michael has a jar containing is dimes and 20 quarters. He randomly draws 3 coins from the jar one after the other without replacement. What is the probability that the second coin he draws is different from the first and the third.
Answer
Calculate \(P(B|AC)\) = \(\frac{\frac{s}{s+t}\cdot\frac{t}{s+t-1}\cdot\frac{t-1}{0.0s+b-2}}{\frac{s}{s+t}\cdot\frac{(t)(t-1)}{(s+t-1)(s+t-2)}}\)
Step 1 ：Let A be the event that first coin is a dime, B be the event that second coin is different from the first, and C be the event that third coin is the same as the second.
Step 2 ：\(P(B|AC)\) = probability that the second coin is different from the first, given that the first is a dime and the third is the same as the second.
Step 3 ：\(P(B|AC) = \frac{P(ACB)}{P(AC)} = \frac{P(ACB)}{P(A)P(C|A)} \)
Step 4 ：Calculate \(P(A)\) = probability that first coin is a dime = \(\frac{s}{s+t}\)
Step 5 ：Calculate \(P(ACB)\) = probability that the first coin is a dime, second is a quarter, and third is a quarter = \(\frac{s}{s+t} \cdot \frac{t}{s+t-1} \cdot \frac{t-1}{s+t-2} \)
Step 6 ：Calculate \(P(C|A)\) = probability that the third coin is the same as the second given that the first coin is a dime = \(\frac{(t)(t-1)}{(s+t-1)(s+t-2)}\)
Step 7 ：Calculate \(P(B|AC)\) = \(\frac{\frac{s}{s+t}\cdot\frac{t}{s+t-1}\cdot\frac{t-1}{0.0s+b-2}}{\frac{s}{s+t}\cdot\frac{(t)(t-1)}{(s+t-1)(s+t-2)}}\)
