Binomial Distribution Calculating

The binomial distribution

A) An Introduction of binomial distribution

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You can use ​

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to work out ​

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of events involving a number of separate stages.

A special case is where each stage has only ​

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, and the probabilities of these two outcomes are the same at each stage (p and 1 - p).

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tree diagrams

probabilities

two outcomes of interest

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Even when there are more than ​

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, they can usually be grouped into ​

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: those which satisfy the event of interest (often called ​

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), and those which do not (​

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).

At any stage, ​

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.

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two basic outcomes

two categories

'success'

'failure'

P(failure) = 1 — P(success)

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For example, if the aim was to count how often a factor of 6 is thrown when

a die is rolled, then throws of any of 1, 2, 3 and 6 will result in 'success'.

So, in this case: P(success) = ​

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and P(failure) = ​

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.

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Some patterns begin to emerge when these probabilities are shown on the branches of a tree diagram and the number of successes is counted.

The number of paths corresponds to the number of different ways you can order the three probabilities, in each case. The l, 3, 3, 1 sequence of the number of paths, for x = 0, 1, 2, 3, is one ofthe rows of Pascal's triangle, as shown on the right.

Match

Match the following

• $$1\times\left(\frac{2}{3}\right)^0\times\left(\frac{1}{3}\right)^3$$

• $$3\times\left(\frac{2}{3}\right)^1\times\left(\frac{1}{3}\right)^2$$

• $$3\times\left(\frac{2}{3}\right)^2\times\left(\frac{1}{3}\right)^1$$

• $$1\times\left(\frac{2}{3}\right)^3\times\left(\frac{1}{3}\right)^0$$

P(x = 0)

P(x = 1)

P(x = 2)

P(x = 3)

• P(x = 0)

• P(x = 1)

• P(x = 2)

• P(x = 3)

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If the tree diagram is extended to six stages (that is, ​

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), there would be ​

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, and it would be cumbersome to work through the whole process like this.

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six throws of the die

64 different paths

However, the coefficients in Pascal's triangle and the patterns seen in the diagram with three stages can be used to write down expressions for the probabilities of getting 0, 1, 2, 3, 4, 5 or 6 successes.

Multiple Choice

When the number of stages (or trials) is large (> 6), it is hard to use Pascal's triangle to find the binomial coefficients.

The number of paths giving r occurrences out of n cases equals the number of ways of choosing r out of n, which is . . .

• $$C_r^n=\frac{n!}{\left(n-r\right)!r!}$$

• $$P_r^n=\frac{n!}{r!}$$

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We need to have values for n, the number of trials, and for p, the probability of ​

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on any one trial, in order for this to make sense, so there is a family of binomial distributions with two ​

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.

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'success'

parameters

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The parameters of the binomial distribution are n (​

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) and p (​

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) on any one trial.

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the number of trials

the probability of a 'success'

Multiple Choice

If $$X\sim B\left(12,\ 0.2\right)$$ , find the probability that $$X=3$$ .

• $$P\left(X=3\right)=C_3^{12}\left(0.2\right)^9\left(0.8\right)^3$$

• $$P\left(X=3\right)=C_3^{12}\left(0.2\right)^3\left(0.8\right)^9$$

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If 25 fair dice are rolled, find the probability that three 6s are seen.

If X = number of 6s seen, then ​

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Multiple Choice

If 25 fair dice are rolled, find the probability that three 6s are seen.

If X = number of 6s seen, then $$X\sim\left(25,\frac{1}{6}\right)$$ .

• $$P\left(X=3\right)=C_3^{25}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{22}$$  

• $$P\left(X=3\right)=C_3^{25}\left(\frac{1}{6}\right)^{22}\left(\frac{5}{6}\right)^3$$  

B) Mean and variance of the binomial distribution

Mean and variance of the binomial distribution

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If $$X\sim B\left(10,\ 0.2\right)$$ , find the mean and variance of $$X$$ .

Given that n = ​

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and p = ​

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, so q =​

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.

Therefore the mean is np = ​

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and the variance is npq = ​

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.

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10

0.2

0.8

2

1.6

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If $$X\sim B\left(80,\ 0.4\right)$$ , find the mean. variance, and the standard deviation of $$X$$ .

Therefore the mean is ​

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and the variance is ​

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, so the standard deviation is ​

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X is a binomial distribution with mean 8 and variance 6.4.

Then, the value of np = ​

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, npq = ​

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, q = ​

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, p = ​

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, and n = ​

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.

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