Count successes in a fixed number of independent trials that each have the same probability. The number of arrangements times the probability of one arrangement gives each term.
Working out E(X) needed a distribution, and so far every distribution had to be built by hand. Some situations hand one over ready made. The coconut shy is won on 0.2 of the goes played, and a book of tickets buys 10 goes. Let X be the number of those 10 goes that win. Every question the stall could ask, about prizes to order or a fair price for a book, comes off the distribution of X. The stall is reset between goes, which is the sort of thing worth checking before trusting any of it.
A binomial model fits when four things are true. There is a fixed number of trials, here 10 goes. Each trial has two outcomes, win or miss. The probability of a win is the same every time, 0.2, because the stall is reset between goes. And the trials are independent, so one result tells you nothing about the next. Check all four before reaching for the model, because the arithmetic works whether or not they hold.
Start with one particular book: win, win, win, then seven misses. The goes are independent, so the probabilities multiply straight along the row: 0.2 × 0.2 × 0.2 × 0.8 × 0.8 × 0.8 × 0.8 × 0.8 × 0.8 × 0.8. That is 0.2³ × 0.8⁷. Working it out, 0.2³ = 0.008 and 0.8⁷ is about 0.2097, so one book like that has probability 0.0016777. Any other book with three wins has the same probability, because the same numbers are multiplied in a different order.
So how many books hold three wins? Choose which 3 of the 10 goes win. The product rule gives 10 × 9 × 8 = 720 ways to pick three in order, but the order does not matter here, and any 3 goes can be ordered in 3 × 2 × 1 = 6 ways. So the real count is 720 / 6 = 120. Then P(X = 3) = 120 × 0.0016777, which is about 0.20. Naming the three winners names the seven losers at the same time, so counting either one is the same job.
Now what breaks it. A drum holds 20 tickets with 5 winners, and three are drawn without being put back. The chance of a winner starts at 5 / 20 and becomes 4 / 19 after a win, so it is not the same every time, and the draws are not independent either. Two of the four conditions fail together, so no binomial applies. With a very large bag the chance barely moves, and then a binomial is still a fair approximation.
A binomial fits when there is a fixed number of trials, two outcomes each, the same probability every time, and independence. Under those conditions one particular arrangement of r successes in n trials has probability p to the r times (1 - p) to the n minus r. Count the arrangements by picking which trials succeed, dividing out the orderings that do not matter, and multiply the two together. Sampling without replacement changes the probability as you go, so it breaks the model.
You priced a spinner go from its distribution last time. The coconut shy is won on 0.2 of the goes played, and a book of tickets buys 10 goes. Dee wants the chance that a book brings home exactly three coconuts, so she can price the books. X is the number of the ten goes that win.
Before any arithmetic, check the conditions. Which one is worth pausing over?
Jobs that lean on this skill. Follow one to see everything it is built on.
24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.