We present problems involving mixture distributions in the context of choosing bowls of balls, as well as related problems involving Bayes’ formula. Problem 1a and Problem 1b are discussed. Problem 2a and Problem 2b are left as exercises.
There are two identical looking bowls. Let’s call them Bowl 1 and Bowl 2. In Bowl 1, there are 1 red ball and 4 white balls. In Bowl 2, there are 4 red balls and 1 white ball. One bowl is selected at random and its identify is kept from you. From the chosen bowl, you randomly select 5 balls (one at a time, putting it back before picking another one). What is the expected number of red balls in the 5 selected balls? What the variance of the number of red balls?
Use the same information in Problem 1a. Suppose there are 3 red balls in the 5 selected balls. What is the probability that the unknown chosen bowl is Bowl 1? What is the probability that the unknown chosen bowl is Bowl 2?
There are three identical looking bowls. Let’s call them Bowl 1, Bowl 2 and Bowl 3. Bowl 1 has 1 red ball and 9 white balls. Bowl 2 has 4 red balls and 6 white balls. Bowl 3 has 6 red balls and 4 white balls. A bowl is chosen according to the following probabilities:
The bowl is chosen so that its identity is kept from you. From the chosen bowl, 5 balls are selected sequentially with replacement. What is the expected number of red balls in the 5 selected balls? What is the variance of the number of red balls?
Use the same information in Problem 2a. Given that there are 4 red balls in the 5 selected balls, what is the probability that the chosen bowl is Bowl i, where ?
Solution – Problem 1a
Problem 1a is a mixture of two binomial distributions and is similar to Problem 1 in the previous post Mixing Binomial Distributions. Let be the number of red balls in the 5 balls chosen from the unknown bowl. The following is the probability function:
The above probability function is the weighted average of two conditional binomial distributions (with equal weights). Thus the mean (first moment) and the second moment of would be the weighted averages of the two same items of the conditional distributions. We have:
See Mixing Binomial Distributions for a more detailed explanation of the calculation.
Solution – Problem 1b
As above, let be the number of red balls in the 5 selected balls. The probability must account for the two bowls. Thus it is obtained by mixing two binomial probabilities:
The following is the conditional probability :
Answers for Problem 2
Let be the number of red balls in the 5 balls chosen random from the unknown bowl.