Binomial Distribution
A binomial distribution occurs where there are multiple trials, where each outcome can be success or failure.
We use combinations, nCk, to count the number of ways to get k successes in n trials.
If the probability of success at an individual trial is p, then the probability of failure is q = 1-p
Formula (to memorise)

- Necessary conditions to use binomial distribution:
- Fixed number of trials, n;
- Only two possible outcomes (success / failure);
- Fixed probability of success, p;
- Trials are independent of each other
(N.B. In exam questions these conditions must be related to the context of the questions).
Worked Example
A regular pentagonal spinner is shown. Find the probability that 10 spins produce exactly three As.

Worked Example 2
- The random variable X is distributed B(12, 1/6). Find:
- P(X=2)
- P(X=9)
- P(X≤1)
Worked Example 3
- The probability that a randomly chosen member of a class is left-handed is 0.15. A random sample of 20 members of the class is taken.
- Suggest a suitable model for the random variable X, the number of members in the sample who are left-handed. Justify your choice.
- Use your model to calculate the probability that:
- Exactly 7 of the members in the sample are left-handed;
- Fewer than 2 of the members in the sample are left-handed.
Exercise



Answers


Expectation and Variance of Binomial Distribution
It is possible to write out the distribution and calculate it as in the previous exercise. Or:
Based on symmetry, we have that E(X) = np and Var(X) = np(1-p)
Worked Example 1
Given that X ~ B(12, 0.3), find the mean, the variance and the standard deviation of X .
Worked Example 2
The random variable X ~B(n, p). Given that E(X ) = 12 and Var(X ) = 7.5, find:
a) the value of n and of p;
b) P(X=11).
Exercise


Answers


Geometric Distribution
Imagine that we are trying to roll a “1” with a fair 6-sided die.
How likely is it that we would succeed on the first time? What about the second time? etc.
P(1 on first roll) = p
P(1 on second roll) = (1-p)p
P(1 on 3rd roll) = (1-p)2p
The following table shows the probabilities of success in this kind of “geometric” distribution:
r | 1 | 2 | 3 | 4 | … | n | … |
P(X=r) | p | p(1-p) | p(1-p)2 | p(1-p)3 | … | p(1-p)4 | … |
Note that the values of P(X=r) in the above table are in geometric sequence with first term p and common ration (1-p), so their sum to infinity is . This makes the geometric distribution suitable as a probability distribution, because the sum of the probabilities is 1.
- Necessary conditions to use geometric distribution:
- Repeated trials are independent of each other;
- Repeated trials can be infinite in number;
- Each trial has exactly two possible outcomes (success / failure);
- Probability of success in each trial, p, is constant.
Formula

Worked Example 1
- Repeated independent trials are carried out in which the probability of success in each trial is 0.66. Correct to 3 significant figures, find the probability that the first success occurs:
- on the third trial;
- on or before the second trial;
- after the third trial.
- In a particular country, 18% of adults wear contact lenses. Adults are randomly selected and interviewed one at a time. Find the probability that the first adult who wears contact lenses is:
- one of the first 15 interviewed;
- not one of the first nine interviewed.
- A coin is biased such that the probability of obtaining heads with each toss is equal to 1/5 . The coin is tossed until the first head is obtained. Find the probability that the coin is tossed:
- at least six times;
- fewer than eight times.
Exercise



Answers

Mode of the geometric distribution
- With all geometric distributions:
- X=1 is the most likely value; and
- P(X=r) decreases as r increases (because 1-p is less than 1 and is being multiplied each time).
Expectation of the geometric distribution
If , then E(X) = 1/p.
Worked Example 1
One in four boxes of cereal contains a free gift. Let the random variable X be the number of boxes that a child opens, up to and including the one in which they find their first gift.
- Find the mode and the expectation of X; and
- Interpret the two values in the context of this question.
Worked Example 2
The random variable X follows a geometric distribution. Given that E(X) = 3½, find P(X>6).
Worked Example 3
Given that and that P(X≤3) = 819/1331, find:
- P(X>3)
- P(1<X≤3).
Exercise


Mixed Exercise


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Mixed Exercise Answers

