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Sunday, July 26, 2020

Probability Part 4 - Notes

Here are the notes for this section.  We will discuss probability using permutations and combinations, the birthday problem, and expected value.  Let me know if you have any questions.

A 4 digit PIN number is selected. What is the probability that there are no repeated digits? Answer: Total possible pin numbers: (10)(10)(10)(10) = 10,000. No repeated digits means that all four digits have to be different and so there are 10 P 4 = 10!/(10 - 4)! choices. Thus, the probability is the # ways for no repeated digits / total # of PIN numbers = 10 P 4 / 10,000 = (10!/(10 - 4)!)/10,000 = 63/125.

Example: In a lottery, 48 balls numbered 1 through 48 are placed in a machine and 6 of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn doesn't matter. Compute the probability that you win the million dollar prize if you purchase a single lottery ticket. Answer: # ways six numbers can be drawn is 48 choose 6 ways. There is only one winner so the probability is 1/(48 choose 6) = 1/12,271,512. Example: In the lottery from the previous example, if 5 of the 6 numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket. Answer: # possible outcomes = 48 choose 6, # ways to choose 5 of the 6 winning numbers is 6 choose 5, # ways to choose 1 out of the 42 losing numbers is 42 choose 1. Thus, the number of ways to win the second prize is given by the Basic Counting Rule: (6 choose 5) times (42 choose 1). So, the probability of winning second prize is (6 choose 5) times (42 choose 1) divided by 48 choose 6 which equals 21/1,022,626


Birthday Problem. Example: Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people? Answer: Probability if at least one birthday in common equals 1 - probability of no birthdays in common. We’ll find P(no shared birthday). Let the three people be denoted by X, Y, and Z. Assume that there are 365 days in a year. The probability that X and Y do not share a birthday is 364/365. The probability that X does not have the same birthday as Y or Z is 363/365. Thus, P(no shared birthday) = (364/365) times (363/365). Hence, P(shared birthday) = 1 - (364/365) which is approximately 0.0082.

Expected Value is the average gain or loss of an event if the procedure is repeated many times. We can compute the expected value by multiplying each outcome by the probability of that outcome, then adding up the products. Example: In a lottery, 48 balls numbered 1 through 48 are placed in a machine and 6 of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. If they match 5 numbers, they will win $1,000. It costs $1 to buy a ticket. Find the expected value. Answer: P(matching all 6 numbers) = 1/12,271,512, P(matching all 5 numbers) = 252/12,271,512. The expected value is ($999,999)(1/12,271,512) + ($999)(252/12,271,512) + (-$1)(12,271,259/12,271,512) is approximately -$0.898. On average, one can expect to lose about 90 cents on a lottery ticket.

If the expected value of a game is 0, we call it a fair game. Example: A 40 yr. old man has a 0.242% risk of not surviving the next year. An insurance company charges $275 for a life insurance policy that pays a $100,000 death benefit. What is the expected value for the person buying the insurance? Answer: The expected value is ($99,275)(0.00242) + (-$275)(0.99758) which is approximately equal to -$33.

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