Suppose there is probability p of occurrence on any one attempt. If we make n independent attempts, then the binomial random variable, denoted by X ~ b(n, p), counts the total number of occurrences in these n attempts.

The **probability** of having exactly k occurrences, for 0 <= k <= n, is given by

The **average** (or **mean**) number of occurrences is given by E[X] = n*p.

The **variance** of the number of occurrences is given by Var(X) = n*p*q, and the **standard deviation** in the number of occurrences is Sqrt(n*p*q).

The most likely number of occurrences, or **mode**, is given by the largest integer k such that k <= (n + 1)*p. The random variable is bimodal when (n + 1)*p is an integer. In that case, this value of k = (n + 1)*p and the previous integer k - 1 will be the modes.

There is no closed-form formula for the **cumulative probability** P(X <= k), or for computing probabilities such as P(j <= X <= k).

The **BINOMIAL** program can be used to compute probabilites such as P(j <= X <= k), P(X = k), and P(X <= k). To execute the program, we enter the values of n and p and the lower and upper bounds of j and k. (Enter the same value k for both the lower and upper bound to compute a pdf value P(X = k).) The program also asks if you want a complete distribution to be entered into the **STAT Edit** screen. If so, then enter **1**. If not, then enter **0**. The program then displays P(j <= X <= k) along with the average value, standard deviation, and mode.

If you enter 1, then the distribution is entered into the **STAT Edit** screen. Under **L1**, the possible number of occurrences 0, 1, . . . , n are listed. Adjacent under **L2**, the pdf values of P(X = k), for 0 <= k <= n, are listed. Under **L3**, the cdf values P(X <= k) are listed.

Click here for info on the TI-86 and TI-89 Stat Edit displays.

If you pressed **1** for a complete distribution, then press **GRAPH** to see a scatterplot of the distribution.

** Example.** Suppose you roll two dice 36 times. On each roll, the probability of getting either a 7 or an 11 is 8 / 36. (a) What is the probability of rolling a 7 or an 11

(a) anywhere from 5 to 10 times in the 36 rolls?

(b) at most 9 times in the 36 rolls?

(c) What is the average number of times and the most likely number of times to roll a 7 or an 11 in 36 rolls?

*Solution.* After calling up the **BINOMIAL** program, enter **36** for **NUMBER OF TRIALS**, enter **8 / 36** for **PROBABILITY**, enter **5** for **LOWER BOUND**, and enter **10** for **UPPER BOUND**. Also enter **1** for a complete distribution.

(a) We see that P(5 <= X <= 10) = 0.769153.

(b) Under list **L3** (**fStat** on the TI-86, or **c3** in **APPS, 6, 1** on the TI-89), we see that P(X <= 9) = 0.73388.

(c) The mean, or average number of rolls, is displayed as 8. The mode, or most likely number of rolls, is also displayed as 8. Under list **L2** (**yStat** on the TI-86, or **c2** in **APPS, 6, 1** on the TI-89), we see that P(X = 8) = 0.15817. So around 15.8% of the time, there will be exactly 8 rolls in 36 tries that give a 7 or an 11.

1. If you flip a coin 15 times, find the probability of getting

(a) exactly 6 heads,

(b) at least 7 heads.

(c) What are the most likely number and the average number of heads in 15 flips?

2. When dealing 5 cards from a shuffled deck, the probability of dealing a
hand with exactly one pair is about 0.42257. In 60 such deals, what is the
probability of having at least 30 hands with one pair?

3. In the 1992 Presidential Election, the independent candidate Ross Perot received a surprising 19% of the popular vote. If an exit pollster
had asked 50 voters at random, what would have been the probability that at
most 15 would have voted for Perot?

1. Here, X ~ b(15, 0.5). In the **BINOMIAL** program, first enter **15**, then **.5**. To find P(X = 6), enter **6** for both the **LOWER BOUND** and **UPPER BOUND**. Also enter **1** for a complete distribution.

(a) We see that P(X = 6) = 0.15274.

(b) Here, P(X >= 7) = 1 - P(X <= 6) = 1 - 0.30362 = 0.69638. (The value for P(X <= 6) is found under list **L3** (**fStat** on the TI-86, or **c3** in **APPS, 6, 1** on the TI-89).

(c) This distribution is bimodal; it is most likely and equally likely that we would have 7 heads or have 8 heads, and from list **L2** (**yStat** or **c2**) we see that P(X = 7) = P(X = 8) = 0.19638, which is the highest occurring probability. But the average number of heads is 7.5

2. Here X ~ b(60, 0.42257) and we wish to find P(X >= 30). In the program, first enter the values **60** and **.42257**. Then enter **30** for **LOWER BOUND** and **60** (the maximum possible number of occurrences) for **UPPER BOUND**. Here we do not need a complete distribution, so enter **0** when asked. We find that P(X >= 30) = 0.1395 . That is, there is only about a 14% chance of dealing at least 30 hands that have exactly one pair when dealing 60 hands independently.

3. Here X ~ b(50, .19) and we wish to find P(X <= 15). In the program, first enter the values **50** and **.19**. Then enter **0** (the minimum possible number of occurrences) for **LOWER BOUND** and **15** for **UPPER BOUND**. We find that P(X <= 15) = 0.9803 .

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