The Binomial Distribution

THE BINOMIAL DISTRIBUTION

1. Factorials

Activity1:Give each member of the class five different coloured multilink cubes. Get them to investigate how many different ways there are of ordering
a) two different coloured cubes,
b) three different coloured cubes,
c) four different coloured cubes,
d) five different coloured cubes.

Suppose we have four different coloured cubes. To work out how many different ways the cubes can be arranged, consider each cube in turn.

The first cube can be any of the four colours. The second cube can be any of the three remaining colours. The third cube can be any of the two remaining colours. Finally, only one colour remains for the final cube. So the total number of ways is

This multiplication is known as ‘4 factorial’, and is written as 4! In general,

Example1:There are six different colours of jelly beans in a bag. In how many ways can they be drawn out? What is the probability of any particular way occurring?

Example2:There are eight runners in a 100m race. Find the number of ways the runners could finish. Explain why we cannot use this to find the probability of any particular way occurring.

Some ways may be more likely than others, such as those ways where the better runners finish before the slower runners.

It is important to be able to manipulate expressions involving factorials.

Example3: Find without a calculator.

Example 4 : Find the number of days in 10! seconds, without using a calculator.

Example5:Simplify

2. Repetition

Suppose the letters of the word PUPIL are jumbled up, and then arranged in a straight line. In how many ways can this be done? If we count the two Ps as being different, by calling them P1 and P2, then there are 5! = 120 ways.

However, there is no visible difference between

and so on, so the number of different arrangements is only .

Now consider the same situation with the letters of the word PUPPY. In particular consider the arrangement P1 U P2 Y P3. Now the three Ps can be arranged in any of 3! = 6 ways in this arrangement, and the result will still always be PUPYP. So the number of different arrangements will be

Finally, consider the letters of the word BANANA. In particular consider the arrangement A1 N1 A2 B A3 N2. The three As can be arranged in 3! = 6 ways within this arrangement, and the two Ns in 2! = 2 ways. So each arrangement is repeated times. Therefore the number of different arrangements will be

In general if there are k different types of objects with of each type, then the number of different arrangements is

Example1:The letters of the word LONDON are jumbled up and then placed in a line.
a) Find the number of different arrangements of the six letters.
b) Find the probability of obtaining the word LONDON.

There are 6 letters : 2 Os, 2 Ns, 1 L and 1 D.

Example2:The letters of the word STATISTICS are jumbled up and then placed in a line.
a) How many different arrangements are there?
b) Find how many of these arrangements start and end with the letter S.

a) There are 10 letters : 3 Ss, 3 Ts, 2 Is, 1 A and 1 C.

b)If an arrangement has to start and end with the letter S, then there 8 letters remaining : namely 1 S, 3 Ts, 2 Is, 1 A and 1 C.

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3. Combinations

The previous section has brought us very close to the idea of a combination. Suppose we choose a word, such as MAMMA, made up of only two different letters. In how many different ways can these letters be arranged? There are five letters, and so there are 5! arrangements, but we must divide by 3! for the three Ms, and by 2! for the two As. So the number of combinations is

In general if we have n letters, with r of one type and of another, then the number of combinations is given by

Example1:How many ways are there of arranging 5 blue cubes and 3 red cubes in a row.

Notice that we could also have used here, as we can think of choosing 5 cubes to be blue as choosing 3 cubes to be red. This illustrates the symmetrical property that

We don’t need to use the formula for , as there it has a calculator function. To find on the Casio fx-9750G, type 8 OPTN F6 (>) F3 (PROB) F3 (nCr) 5 EXE

Example2:A swimming team of five is to be selected from a squad of 7. How many possible teams are there?

This situation is analogous to arranging letters or cubes, because we must order five ticks and two crosses for swimmers A to G. One possible order would be

Example3:How many different sets of six numbers can I choose on my National Lottery ticket?

Once again this questions is effectively asking us to order a number of ticks (6) and a number of crosses (43).

Example4:There are five maths teachers and six english teachers in a meeting. How many ways are there of choosing a subcommittee of two maths teachers and three english teachers?

The maths teachers can be chosen in ways, and the english teachers can be chosen in ways. Therefore

Example5:This time the subcommittee of five is chosen by drawing names from a hat. What is the probability there are no maths teachers on the subcommittee?

To have no maths teachers we must choose five english teachers from the six available.

We could also find the answer like this, without using combination theory.

Example6:There are 7 staff and 6 students on the sports council of a college. A committee of 8 people from the 13 on the council is to be selected to organise a tennis competition. Find the number of different ways in which the committee can be selected if there must be more students than staff.

We can have 5 students and 3 staff, or 6 students and 2 staff.

To select 5 students from 6 and then 3 staff from 7, we must do .

Similarly, to select 6 students from 6 and then 2 staff from 7, we must do .

Example7:A hockey team of 11 players is to be selected from a group of 20 students. Three of the students can play in goal, the others can only play in any of the remaining positions. How many different teams are possible?

We must choose 1 goalkeeper from 3, and then 10 other players from 17.

4. The Binomial Distribution

As a keen gardener, I plant a lot of seeds. Yesterday, I planted three seeds. The packet said that each seed has a 0.8 probability of producing flowers. What are the probabilities of obtaining 2 flowers?

At the moment, the only way we have of answering this question is to draw the tree diagram opposite, which takes ages.

We can see that FFN has a probability of . But there is also FNF and NFF that give two successes - in other words there are three ways, all with a probability of 0.128, because they all involve multiplying 0.8 by 0.8 by 0.2, albeit in different orders.

So the total probability is .

We can summarise the results in a table.

There are two things to notice here.

•The probability of each way of getting n flowers is , because we need n seeds to produce flowers, and seeds not to.

•The number of ways (1, 3, 3, 1) are actually combinations, from our previous section. For example, to get 2 flowers, we need all the ways of arranging two Fs and one N, which is given by.

Example1:The next day, I plant four seeds. Find the probability that exactly half of them produce flowers.

Example2:The day after I plant ten seeds. What is the probability that

a) seven seeds produce flowers?

b) four seeds produce flowers?

c) less than 9 seeds produce flowers?

Clearly we do not want to draw a ten level tree diagram, as this will have different outcomes!

Example3: Long experience tells me that my favourite chat-up line ‘OK darling, get your coat - you’ve pulled’ works with probability . What is the probability it will work exactly twice in the next five attempts?

Example 4:The probability that a girl is left-handed is 0.15. A group of 10 girls is selected at random. Find the probability that…

a)there are two left-handed girls in the group

b)there is one left-handed girl in the group

c)there are more than two-left handed girls in the group.

Consider the table of probabilities at the start of this section.

As in unit S1, this is known as a probability distribution. More technically, it is known as a binomial distribution, and can be written as follows.

The list below gives the conditions necessary for the use of a binomial distribution.

•There is a fixed number (n) of trials. In example 2 of section 1 there were 10 trials (10 seeds planted). In example 3, there were 5 trials (5 chat-ups).

•The trials must be independent. This means that the probability of success p does not change with each successive trial. This may not strictly be the case for our examples, as I could improve my horticultural or chatting-up skills with experience.

•The trials have only two outcomes - success or failure. From the Greek, bi = two, nomen = terms. Examples include a child being a boy or a girl, a pupil in my class either listening to me or not listening to me, etc. etc.

If these conditions are fulfilled, then we have the binomial probability function

So in example 2, X represents the number of seeds producing flowers and

Similarly, in example 3, X represents the number of successful chat-ups and

If X is a binomial distribution with n trials, and probability of trial success p, we use the notation

So our examples have binomial distributions and .

Example5:Every week I order a takeaway. The probability it’s a curry is 0.7, and the probability it’s a Chinese is 0.3. Find the probability that in the next five weeks I’ll have exactly three curries.

Activity2:At this stage it is probably worth using Autograph to illustrate the binomial distribution. Open a statistics page, click on the probability distribution icon , select binomial, and enter the parameters and . Set the axes to suitable scales. Click on the probability calculations icon and enter 3 for lower r and 3 for upper r. The answer 30.87% will appear at the bottom of the screen.

Example6:My dartboard has a picture of Tony Blair on it. I find I can hit him on the nose with a dart with probability 0·2. What is the probability I hit him four times out of six?

Example7:I throw a fair dice four times. Find the probability I get exactly one six.

Example8:Every morning my alarm goes off at 7 a.m. The probability I get up immediately is . Find the probability I get up immediately at least five times a week.

Example9:The probability I will add up the marks of an exam correctly is 65%. Find the probability I will make no more than one error when adding up the marks of ten exams.

Example10:Seven fair coins are tossed, and those landing heads up are eliminated. The remainder are tossed again, and the process of elimination is repeated. This continues until no coins are left. Find the probability that
a)exactly three coins are eliminated in the first round,

b)exactly three coins are eliminated in the second round, given that exactly three coins are eliminated in the first round,

c)exactly three coins are eliminated in the first round, and exactly three coins are eliminated in the second round,

d)exactly three coins are eliminated in the first round, and exactly three coins are eliminated in the third round.

d) Let Z be the number of coins eliminated in the third round. We require either of the following.

Activity3:Use Autograph to plot the binomial distribution with to simulate the number of heads when four coins are tossed. Then click the probability animation icon and animate p from 0 to 1 in steps of 0.01. Now try animating n instead. Comment on the distributions.

5. Binomial Cumulative Probability Tables

These tables can be found on p121 of the text book and in much more detail on p15-19 of the formula book. They give the probability of for for various values of x, n and p. This saves the drudgery of adding together lots of individual probabilities.

Example1:The probability a light bulb will fail in under 100 hours of use is 0·2. A new house is fitted with 20 bulbs. Use cumulative binomial probability tables to find the probability that

a) no more than 6 fail in under 100 hours of use,

b) at least 3 fail in under 100 hours of use,

c) exactly four fail in under 100 hours of use.

Checking part c) with our other method,

Activity4:Autograph can be used to illustrate the above results (and for the rest of the questions in this section). Plot the binomial distribution with parameters . Then click on the probability calculations icon and enter the following for the relevant part of the question.
a) cumulative b) cumulative c) lower r = 4, upper r = 4.
The probabilities will appear at the bottom of the screen.

Example2:A coin is biased so that the probability of it landing ‘heads up’ is . Find the probability there will be less than six heads in the next ten throws.

The tables only deal with probabilities less than or equal to 0.5. So we must consider tails, rather than heads. The number of tails X follows the distribution

Activity5:Use Excel to produce a table for , and use it to check the answer above. The COMBIN function generates binomial coefficients.

Copy columns B and C down as far as row 12.

Example3:I throw water balloons at passers-by from my upstairs window. The probability I hit any particular person is 0·4. One evening, fifteen people pass me by. Find the probability I hit between six and nine people inclusive.

Example4:The probability that I will get home in time for ‘Neighbours’ is 95%. Find the probability that I will watch between 42 and 45 inclusive of the next 50 episodes.

We need to consider the number of times X I miss this fine antipodean drama, which has a probability of 0.05.

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6.The Mean and Variance of a Binomial Distribution

If the probability of a football player being sent off in any particular match is , then in 20 matches we would expect him to be sent off times. In general the expectation is the number of trials multiplied by the probability of success. The formula for the variance is also straightforward, and is quoted below – you need not worry about where it comes from!

Example1:A fair dice is rolled 10 times. What is the most likely number of sixes to come up? Find the probability of this.

This is impossible so we round to the nearest integer. The most likely number of sixes is two.

Example2:It is estimated that 1 in 20 people are left-handed.

a)What size sample should be taken to ensure that the expected number of left-handed people is 3?

b)What is the probability of having exactly three left-handed people in this case?

The above examples show that although the expectation is the most likely number of successes, it may well be unlikely to occur!

Example3: The probability that an apple, picked at random from a sack, is bad is 0.15. For a sack of 40 apples,
a) Find the expected number of bad apples.
b) Find the standard deviation in the number of bad apples.

c)Find the probability that the number of bad apples is within one standard deviation either side of the mean.

Example4:The probability that a pupil at a particular school will pass GCSE Maths is 0.7. If 30 pupils are entered for the examination, find the expected number of passes, and the standard deviation.
If the number of passes is more than one standard deviation above the mean, the Head of Maths gets a salary bonus. Find the probability this will happen.

As our tables only deal with probabilities less than or equal to 0.5, we need to consider the probability that the number of students not passing Y is less than or equal to 6.

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