Probability and Binomial Distributions
Finding the Distribution

You will have already met this type of distribution in earlier sections of this unit, as well as the concept of random variables.

Worked Examples

1

Ashoke, Theo and Sadie will each visit the local leisure centre to swim on one evening next week but have made no arrangement between themselves to meet or go on any particular day. The random variable X is 'the number of the three who go to the leisure centre on Wednesday'. Find the probability distribution for X, that is, the probabilities for each value of X.

Show me

The probabilities of 0, 1, 2 or 3 people going on Wednesday can be found by using the tree diagram method covered in the unit Probability of Two or More Events.

The following tree diagram shows probabilities for how many go on Wednesday.

You can see that

P(X = 3) = \left(\frac{1}{7}\right)^3,   P(X = 0) = \left(\frac{6}{7}\right)^3

and

P(X = 2)= \frac{1}{7}\times\frac{1}{7}\times\frac{6}{7}+\frac{1}{7}\times\frac{6}{7}\times\frac{1}{7}+\frac{6}{7}\times\frac{1}{7}\times\frac{1}{7}
= 3\times\left(\frac{1}{7}\right)^2\left(\frac{6}{7}\right)
P(X = 1)= \frac{1}{7}\times\frac{6}{7}\times\frac{6}{7}+\frac{6}{7}\times\frac{1}{7}\times\frac{6}{7}+\frac{6}{7}\times\frac{6}{7}\times\frac{1}{7}
= 3\times\left(\frac{1}{7}\right)\left(\frac{6}{7}\right)^2

This gives the table below:

x0123
P(X = x)\left(\frac{6}{7}\right)^33\left(\frac{1}{7}\right)\left(\frac{6}{7}\right)^23\left(\frac{1}{7}\right)^2\left(\frac{6}{7}\right)\left(\frac{1}{7}\right)^3
=\frac{216}{343}=\frac{108}{343}=\frac{18}{343}=\frac{1}{343}

The method used in the example above can be extended to a fourth person so that there will be sixteen branches to cover all the possibilities as shown in the diagram below, with X now 'the number of the four people who go on Wednesday'.

The resulting probability distribution is

x01234
P(X = x)\left(\frac{6}{7}\right)^44\left(\frac{1}{7}\right)\left(\frac{6}{7}\right)^36\left(\frac{1}{7}\right)^2\left(\frac{6}{7}\right)^24\left(\frac{1}{7}\right)^3\left(\frac{6}{7}\right)\left(\frac{1}{7}\right)^4

The fractions are as you might expect. For instance, looking at

P(X = 2) = 6\left(\frac{1}{7}\right)^2\left(\frac{6}{7}\right)^2

since you are interested in having two present and the probability for each is \frac{1}{7}, the \left(\frac{1}{7}\right)^2 is explained, and as two are not to attend this produces the \left(\frac{6}{7}\right)^2.

The multiplier '6' has come from the second of the tree diagrams: there are six branches corresponding to two being present. This is because there are six ways of writing down two \left(\frac{1}{7}\right)s and two \left(\frac{6}{7}\right)s in a row and each produces a branch.

The six ways are shown below:

\frac{1}{7} × \frac{1}{7} × \frac{6}{7} × \frac{6}{7}, \frac{1}{7} × \frac{6}{7} × \frac{1}{7} × \frac{6}{7}, \frac{1}{7} × \frac{6}{7} × \frac{6}{7} × \frac{1}{7},
\frac{6}{7} × \frac{1}{7} × \frac{1}{7} × \frac{6}{7}, \frac{6}{7} × \frac{1}{7} × \frac{6}{7} × \frac{1}{7}, \frac{6}{7} × \frac{6}{7} × \frac{1}{7} × \frac{1}{7}

There are four fractions to write down, two of each type, and the number of different ways of combining these is six. You don't want to draw a tree diagram every time so another method can be developed. For example, having ten people going to the leisure centre would need a tree with 2^{10} = 1024\left(\frac{1}{7}\right)^3 branches.

If you can produce the fractions in the probability distribution then all that is needed is a way of getting the right numbers to put with them. With three people the numbers were 1, 3, 3, 1 and with four they were 1, 4, 6, 4, 1.

You might recognise these as being rows of Pascal's Triangle, shown below.

For three to be present you need \left(\frac{1}{7}\right)^3 and for seven to stay away the term is \left(\frac{6}{7}\right)^7.

When writing down these fractions, how many different ways are there of combining them? According to Pascal's Triangle there are 120, so

P(X = 3) = 120\left(\frac{1}{7}\right)^3\left(\frac{6}{7}\right)^7

which is easier to calculate than drawing 1024 branches! The number of ways of choosing 3 from 10 is often written

\binom{10}{3}   or   ^{10}\text{C}_3

so

P(X = 3) = \binom{10}{3}\left(\frac{1}{7}\right)^3\left(\frac{6}{7}\right)^7

and the probability distribution is

P(X = x) = \binom{10}{x}\left(\frac{1}{7}\right)^x\left(\frac{6}{7}\right)^{10-x}\ \ \ x = 0, 1, ..., 10

Whilst the values needed can easily be read off Pascal's Triangle, there is an even easier way of working out the coefficients given in terms of factorials. Note that n! = n × (n - 1) × ... × 2 × 1 and, for example,

\binom{10}{3}= \frac{10!}{7!3!}
= \frac{10 × 9 × 8 × (7 × 6 × 5 × 4 × 3 × 2 × 1)}{(7 × 6 × 5 × 4 × 3 × 2 × 1) × (3 × 2 × 1)}
= \frac{10 × 9 × 8}{3 × 2 × 1}\ \ \left(\text{since} \frac{10!}{7!} = 10 × 9 × 8\right)
= 120
2

Check the values of

\binom{10}{1}, \binom{10}{2}, \binom{10}{4} and \binom{10}{5}

by using Pascal's Triangle and the factorial formula.

Show me

\binom{10}{1} = \frac{10!}{9! 1!} = 10

\binom{10}{2} = \frac{10!}{8! 2!} = \frac{10 × 9}{2} = 45

\binom{10}{4} = \frac{10!}{6! 4!} = \frac{10 × 9 × 8 × 7}{4 ×3× 2 ×1} = 210

\binom{10}{5} = \frac{10!}{5! 5!} = \frac{10 × 9 × 8 × 7 × 6}{5× 4 ×3× 2 ×1} = 252

It should also be noted that most calculators have the facility to produce the coefficients.
Another notation used for the number of different ways of combining three from ten is ^{10}\text{C}_3, and the majority of calculators have a key labelled with one of ^{n}\text{C}_r or \binom{n}{r}
The instructions for your calculator will show you how to use this function.

In general, when the probability of success is p (instead of ), and the experiment is repeated n independent times with the probability remaining constant, the probability distribution for the number of successes is given by

P(X = r) = \binom{n}{r}p^r(1 - p)^{n-r}   r = 0, 1, ..., n

The notation X ~ B (n, p) is often used.

A practical experiment illustrates its use.

Take five cards numbered from one to five. Seat two people back-to-back and give the cards to one of them. This person selects a card at random and the other participant tries to identify it.

This is done five times and repeated with other pairs of people and you record the result. You can use the Binomial distribution to find the expected number of successes, assuming no skills beyond guessing!

So, if X is 'the number predicted correctly out of five attempts', the probability distribution is

P(X = r) = \binom{5}{1}(0.2)^r(0.8)^{5-r}   for x = 0, 1, ... 5

So, for example, the probability of getting one correct is given by

P(X = 1)= \binom{5}{1}(0.2)(0.8)^4
= 5(0.2)(0.8)^4
= 0.4096

Hence for 20 people, you would expect

20 × 0.4096 = 8.192 ≈ 8

to get one correct.

A table of expected frequencies is shown opposite.

Compare your observed frequencies with the theoretical ones.

3

If X is binomially distributed with 6 trials and a probability of success equal to at each attempt, what is the probability of:

(a)   exactly 4 successes (b)   at least one success?

Show me

This question can be rewritten in the following way.

If X ~ B \left(6,\frac{1}{4}\right), what is: (a)   P(X = 4) (b)   P(X ≥ 1) ?

(a)

P(X = 4)= \binom{6}{4} \binom{1}{4}^4 \binom{3}{4}^2
= 15 × \frac{1}{256} × \frac{9}{16}
= \frac{135}{4096} = 0.033   (to 3 d.p.)

(b)

P(X ≥ 1)= 1 - P(X = 0)
= 1 - \left(\frac{3}{4}\right)^6
= 1 - \frac{729}{4096}
= \frac{3367}{4096} = 0.822   (to 3 d.p.)

Note

Tables giving cumulative binomial probabilities are given in some calculators or using an online search.

4

When an unbiased coin is tossed eight times what is the probability of obtaining:

(a)   less than 4 heads (b)   more than five heads?

Show me

H = number of heads   ⇒ H ~ B(8, 0.5)

(a)

Using the appropriate Binomial Distribution table you can simply write down the answer,

P(H < 4)= P(H ≤ 3)
= 0.3633

Alternatively,

P(H ≤ 3)= P(H = 0) + P(H = 1) + P(H = 2) + P(H = 3)
= \left(\frac{1}{2}\right)^8 + \binom{8}{1}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^7 + \binom{8}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6 + \binom{8}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^5
= \left(\frac{1}{2}\right)^8+8\left(\frac{1}{2}\right)^8+28\left(\frac{1}{2}\right)^8+56\left(\frac{1}{2}\right)^8
= 93\left(\frac{1}{2}\right)^8
= \frac{93}{256}
= 0.3633   (to 4 d.p.)

(b)

P(H > 5)= 1 - P(H ≤ 5)
= 1 - 0.8555   (from tables)
= 0.1445

or

P(H > 5)= P(H = 6) + P(H = 7) + P(H = 8)
= \binom{8}{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2 + \binom{8}{7}\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^8
= 28\left(\frac{1}{2}\right)^8 + 8\left(\frac{1}{2}\right)^8 + \left(\frac{1}{2}\right)^8
= 37\left(\frac{1}{2}\right)^8
= \frac{37}{256}
= 0.1445   (to 4 d.p.)

Exercises

Where decimals are used give answers correct to 3 significant figures.

If X ~ B(6, ) find:

(a)
P(X = 2)
(b)
P(X < 2)
(c)
P(X ≥ 1)

If X ~ B(10, 0.3) find:

(a)
P(X = 9)
(b)
P(X = 0)
(c)
P(X ≤ 5)

A regular tetrahedron has three white faces and one red face. It is rolled four times and the colour of the bottom face is noted. What is the most likely number of times that the red face will end downwards?

If the probability that I get a lift to work on any morning is 0.6 what is the probability that in a working week of five days I will get a lift only twice?

When a consignment of pens arrives at the retailer's, ten of them are tested. The whole batch is returned to the wholesaler if more than one of those selected is found to be faulty. What is the probability that the consignment will be accepted if 2% of the pens are faulty?

The probability that a certain type of vacuum tube will shatter during a thermal shock test is 0.15.

(a)

What is the probability that if 25 such tubes are tested

(i)

4 or more will shatter,

(ii)

between 16 and 20 (inclusive) will survive?

(b)

Another type of tube is tested in samples of 30. It is observed that on 40% of occasions all 30 survive the test. What is the probability (assumed constant) of a single tube of this type surviving the test?

A crossword puzzle is published in The Times each day of the week, except Sunday. A man is able to complete, on average, 8 out of 10 of the crossword puzzles.

(a)

Find the expected value and the standard deviation of the number of completed crosswords in a given week.

expected value:

standard deviation:

(b)

Given that he completes the puzzle on Monday, find, to 3 significant figures, the probability that he will complete at least 4 in the rest of the week.

(c)

Find, to 3 significant figures, the probability that, in a period of four weeks, he completes 4 or less in only one of the four weeks.