Binomial Distributions Probability, Part 2 Binomial Distributions Consider a case where you flip a coin 5 times and you want to know the probability of getting heads 3 times. This is called a binomial probability because each trial has 2 possible outcomes—Heads or Tails.

Binomial Distributions Probability, Part 2 Binomial Distributions We need to first determine the number of possible outcomes for flipping a coin 5 times. We could actually write out all of them: HHHHH HHHHT HHHTH HHHTT HHTHH HHTHT HHTTH HHTTT HTHHH HTHHT HTHTH HTHTT HTTHH HTTHT HTTTH HTTTT THHHH THHHT THHTH THHTT THTHH THTHT THTTH THTTT TTHHH TTHHT TTHTH TTHTT TTTHH TTTHT TTTTH TTTTT

Binomial Distributions Probability, Part 2 Binomial Distributions Now consider how many of these arrangements contain 3 Heads. There are 10 such arrangements. HHHHH HHHHT HHHTH HHHTT HHTHH HHTHT HHTTH HHTTT HTHHH HTHHT HTHTH HTHTT HTTHH HTTHT HTTTH HTTTT THHHH THHHT THHTH THHTT THTHH THTHT THTTH THTTT TTHHH TTHHT TTHTH TTHTT TTTHH TTTHT TTTTH TTTTT

Binomial Distributions Probability, Part 2 Binomial Distributions Of course we really didn’t need to go to all that trouble. Recall from our previous examples with the True/False quizzes (which are also binomial probabilities) that we can simply determine the number of combinations of 3 items out of a list of 5: C(5, 3) = 10

Binomial Distributions Probability, Part 2 Binomial Distributions Next we consider the total possible outcomes for our scenario, which is 32: HHHHH HHHHT HHHTH HHHTT HHTHH HHTHT HHTTH HHTTT HTHHH HTHHT HTHTH HTHTT HTTHH HTTHT HTTTH HTTTT THHHH THHHT THHTH THHTT THTHH THTHT THTTH THTTT TTHHH TTHHT TTHTH TTHTT TTTHH TTTHT TTTTH TTTTT

Binomial Distributions We could then conclude that Probability, Part 2 Binomial Distributions We could then conclude that P(3 Heads) = 10/32 = 0.3125

Binomial Distributions Probability, Part 2 Binomial Distributions We could also view this problem in a slightly different manner. Begin again with the fact that there are 10 possible outcomes that give us 3 Heads: C(5, 3) = 10

Binomial Distributions Probability, Part 2 Binomial Distributions Now look at the probability of getting Heads on any 1 toss, as well as Tails: P(Heads) = ½ = 0.5 P(Tails) = ½ = 0.5

Binomial Distributions Probability, Part 2 Binomial Distributions Because we are looking for 3 Heads, the probability of getting Heads will show up 3 times: (1/2) * (1/2) * (1/2) = (1/2)3

Binomial Distributions Probability, Part 2 Binomial Distributions Because we are looking for 2 Tails, the probability of getting Tails will show up 2 times: (1/2) * (1/2) = (1/2)2

Binomial Distributions Probability, Part 2 Binomial Distributions The probability for getting 3 Heads and 2 Tails is then: (1/2)3 * (1/2)2 = (1/2)5

Binomial Distributions Probability, Part 2 Binomial Distributions Important Note: The strategy just shown should only be used when the events under consideration are independent. Coin flips are independent because the probability of getting heads on one toss is independent of the outcomes of previous tosses; it is always 0.5.

Binomial Distributions Probability, Part 2 Binomial Distributions We could also make an adjustment for a weighted coin. Say the probability of getting Heads were 0.6, rather than the normal 0.5. Then the probability of getting 3 Heads and 2 Tails from 5 tosses would be: P(3 Heads) = 10(0.6)3 (0.4)2 = 0.3456 2 Tails 3 Heads

Probability, Part 2 Expectation Next we want to consider a topic related to probability called expectation. This measure tells us how we should expect to do in the long term, given the probability for an event occurring in a certain way.

Probability, Part 2 Expectation For example, say you are playing a game where you pay $2 to play. If you lose the game, you lose your $2. If you win the game, you win $5.

Now say the probability of winning the game is 0.25. Probability, Part 2 Expectation Now say the probability of winning the game is 0.25. We would expect, then, to win $5, 25% of the time and to lose $2, 75% of the time.

Expectation Our expectation would be .25(5) + .75(-2) = -.25 Probability, Part 2 Expectation Our expectation would be .25(5) + .75(-2) = -.25 P(Win) P(Lose) Win $5 Lose $2

Our expectation would be Probability, Part 2 Expectation Our expectation would be .25(5) + .75(-2) = -.25 In other words, over the long term (perhaps thousands of games) we would expect an average loss of 25¢ per game, or $25 per 100 games.

Probability, Part 2 Example 1: A student completely guesses on a 20 question True/False quiz. What is the probability that the student will answer 10 questions True and 10 False? First, there are C(20, 10) = 184,756 ways to have 10 True responses. Next, P(True) = P(False) = 0.5 So P(10 True) = 184,756(0.5)20 = 0.176.

Probability, Part 2 Example 2: Say that 0.1% of all people are carriers for a certain disease. As carriers, they have one normal gene, N, and one gene, D, which codes for the disease. If two parents are both carriers, what is the probability that their first child will have the disease? (The child would need to receive the D gene from each parent to have the disease.)

Probability, Part 2 Example 2: Because each parent is a carrier, the probability of contributing the D gene is 0.5 for each of them. For the child, then, P(Disease) = (0.5)2 = 0.25.

Probability, Part 2 Example 3: What is the probability that a child will inherit the disease from a couple chosen at random, given that neither parent has the disease?

Probability, Part 2 Example 3: In this case, we don’t know if the parents are carriers. Therefore, the probability for being a carrier is 0.001 for each. For each parent that is a carrier, there will be a 0.5 probability of passing on the D gene. In order to have the disease, the child must receive the D gene from both parents.

Example 3: P(Disease) = (0.001)2(0.5)2 = 0.00000025 Probability, Part 2 Example 3: P(Disease) = (0.001)2(0.5)2 = 0.00000025

Probability, Part 2 Example 4: A company produces video games, and has determined that 10% of them finish the manufacturing process with some type of defect or another. If a quality control inspector looks at 5 games before they leave the factory, what is the probability that at least 1 will be defective?

Probability, Part 2 Example 4: This is another case of binomial probability because each game fits one of two categories: defective, or non-defective.

Probability, Part 2 Example 4: We can attack the problem from two different angles. Because we want the probability of finding at least 1 defective game, we could add together the probabilities of finding 1, 2, 3, 4, and 5 defective games.

Probability, Part 2 Example 4: For 1 defective game: The number of ways to find 1 game out of 5 is C(5, 1). The probability of 1 game being defective is (0.1)1 The probability of 4 games being non-defective is (0.9)4 P(1 defective game) = C(5, 1)(0.1)1(0.9)4

Probability, Part 2 Example 4: P(1 defective) = C(5, 1)(0.1)1(0.9)4 = 0.328 P(2 defective) = C(5, 2)(0.1)2(0.9)3 = 0.073 P(3 defective) = C(5, 3)(0.1)3(0.9)2 = 0.008 P(4 defective) = C(5, 4)(0.1)4(0.9)1 = 0.000 P(5 defective) = C(5, 5)(0.1)5(0.9)0 = 0.000 (Probabilities have been rounded to the nearest thousandth.)

Probability, Part 2 Example 4: P(1 defective) = C(5, 1)(0.1)1(0.9)4 = 0.328 P(2 defective) = C(5, 2)(0.1)2(0.9)3 = 0.073 P(3 defective) = C(5, 3)(0.1)3(0.9)2 = 0.008 P(4 defective) = C(5, 4)(0.1)4(0.9)1 = 0.000 P(5 defective) = C(5, 5)(0.1)5(0.9)0 = 0.000 P(at least 1 defective) = 0.328 + 0.073 + 0.008 + 0.000 + 0.000 = 0.409, or about 41%.

Probability, Part 2 Example 4: We could have taken a different approach, considering instead the probability that none of the 5 games would be found defective: P(0 defective) = C(5, 0)(0.1)0(0.9)5 = 0.590, or 59%.

Probability, Part 2 Example 4: Saying we have a 59% probability of finding no games out of 5 as defective is the same as saying we have a 41% probability of finding at least 1 game out of 5 as defective.