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 .
Return to Table of Contents.