Probability Involving a Weighted Coin A coin is weighted so that the probability of heads is 0.6. What is the probability of getting exactly two heads in five tosses of this coin?
Probability Involving a Weighted Coin Since the tosses are independent, the probability of getting two heads followed by three tails is
Probability Involving a Weighted Coin But this is not the only way we can get exactly two heads. The two heads could occur, for example, on the second toss and the last toss.
Probability Involving a Weighted Coin In this case, the probability is
Probability Involving a Weighted Coin In fact, the two heads could occur on any two of the five tosses. Thus, there are C(5, 2) ways in which this can happen, each with probability (0.6) 2 (0.4) 3. It follows that
Binomial Experiments The probabilities that we have just calculated are examples of binomial probabilities. In general, a binomial experiment is one in which there are two outcomes, which we call “success” and “failure”. In the coin-tossing experiment described previously, “success” is getting “heads” and “failure” is getting “tails”. The following tells us how to calculate the probabilities associated with binomial experiments when we perform them many times.
Binomial Probability An experiment has two possible outcomes. S and F – called “success” and “failure” With P(S) = p and P(F) = 1 – p. The probability of getting exactly r successes in n independent trials of the experiment is P(r successes in n trials) = C(n, r)p r (1 – p) n – r
Binomial Probability and Binomial Coefficient The name “binomial probability” is appropriate because C(n, r) is the same as the binomial coefficient.
E.g. 1—Binomial Probability A fair die is rolled 10 times. Find the probability of each event. a)Exactly 2 sixes. b)At most 1 six. c)At least 2 sixes.
We interpret “success” as getting a six and “failure” as not getting a six. So, P(S) = 1/6 and P(F) = 5/6. Since each roll of the die is independent from the others, we can use the formula for binomial probability with n = 10, p = 1/6. E.g. 1—Binomial Probability
Using these values in the formula gives us: P(exactly 2 are sixes) = C(10, 2)(1/6) 2 (5/6) 8 ≈ 0.29 Example (a) E.g. 1—Binomial Probability
The statement “at most 1 six” means 0 sixes or 1 six. So So P(at most one six) = P(0 sixes or 1 six) = P(0 sixes) + P(1 six) = C(10, 0)(1/6) 0 (5/6) 10 + C(10, 1)(1/6) 1 (5/6) 9 ≈ 0.1615 + 0.3230 ≈ 0.4845 Example (b) E.g. 1—Binomial Probability
The statement “at least two sixes” means two or more sixes. Adding the probabilities that 2, 3, 4, 5, 6, 7, 8, 9, or 10 are sixes is a lot of work. It’s easier to find the probability of the complement of this event. –The complement of “two or more are sixes” is “0 or 1 are sixes”. Example (c) E.g. 1—Binomial Probability
So P(two or more sixes) = 1 – P(0 or 1 six) = 1 – 0.4845 = 0.5155 Example (c) E.g. 1—Binomial Probability
We can describe how the probabilities of an experiment are “distributed” among all the outcomes of an experiment by making a table of values. The function that assigns to each outcome its corresponding probability is called a probability distribution. A bar graph of a probability distribution in which the width of each bar is 1 is called a probability histogram. The next example illustrates these concepts.
E.g. 4—A Binomial Distribution A fair coin is tossed eight times, and the number of heads is observed. Make a table of the probability distribution, and draw a histogram. What is the number of heads that is most likely to show up?
E.g. 4—A Binomial Distribution This is a binomial experiment with n = 8 and p= ½, so 1 – p = ½ as well. We need to calculate the probability of getting 0 heads, 1 head, 2 heads, 3 heads, and so on.
E.g. 4—A Binomial Distribution To calculate the probability of 3 heads, we have
E.g. 4—A Binomial Distribution The other entries in the following table are calculated similarly.
E.g. 4—A Binomial Distribution We draw the histogram by making a bar for each outcome with width 1 and height equal to the corresponding probability.
The Binomial Distribution Notice that the sum of the probabilities in a probability distribution is 1. because the sum is the probability of the occurrence of any outcome in the sample space (this is the certain event).