The binomial distribution is an example of a discrete random variable. It has two parameters n (number of trials) and p (probability of success of one trial): X~B(n , p). For a situation to be described using a binomial model, the following must be true
- A finite number of trials, n are carried out
- The trials are independent
- The outcome of each trial is deemed either a success or a failure
- The probability, p, of a successful outcome is the same for each trial
On this page, you should learn about
- the binomial distribution
- mean of the binomial distribution
- variance of the binomial distribution
- the formula for calculating probabilities
P(X = r) = nCr x pr x (1 - p)n - r
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Here is a quiz that practises the skills from this page
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For a binomial distribution, some of the following statements are TRUE and some are FALSE.
Type T (TRUE) or F (FALSE) next to each one
The probability of one trial depends on the outcome of the previous trial
A finite number of trials is carried out
The distribution is always symmetrical
The binomial distribution is a continuous distribution
The outcome of each trial is deemed either a success or a failure
The mean and variance of the distribution are always equal
The probability of a successful outcome is the same for each trial
The trials are independent
The graph below shows a binomial distribution for the variable X. Answer the probability questions below
P(X = 2) =
P(X = 0) =
P( \(1\le X<4\) ) =
P(X > 0) =
P(X > a) = 0.6875 , a =
P(1 \(\le\) X < 4) = P(X=1) + P(X=2) + P(X=3) = 0.25 + 0.375 + 0.25 = 0.875
The probability of a head on a biased coin is 0.55.
The coin is flipped 8 times. The probability of getting 6 heads is aC6 x bd x 0.45e
Find the values of a , b , d and e
a =
b =
d =
e =
A binomial distribution, X has probability distribution X~B(10,0.3)
P(X = 2) = aCb x 0.3d x e8
Find the values of a , b , d and e
a =
b =
d =
e =
b can also take the value 8 since 10C2 = 10C8
The random variable X has a binomial distribution with n = 12 and p = 0.35.
Calculate the following to 3 significant figures.
- P(X = 3) =
- P(X \(\le\) 3) =
- P(X < 3) =
- P(X \(\ge\) 9) =
- P(2\(\le\) X \(\le\) 10) =
In the ma
ss production of mobile phones, it is found that 5% are faulty. Phones are selected at random and put into boxes of 20. A box is selected at random.
Giving your answers to 3 significant figures find the probability that it will contain
- Three faulty phones
- No faulty phones
- At least two faulty phones
- Less than 2 faulty phones
- The mean number of faulty mobile phones
X~B(20,0.05)
- P(X=3)
- P(X=0)
- P(X \(\ge\) 2)
- P(X \(\le\) 1)
- E(X) = np = 20x0.05
If X~B(n,0.5) and P(X<1) \(\approx\) 0.00391.
Find n
n =
P(X=0) \(\approx\) 0.00391
nC0 x 0.50 x 0.5n \(\approx\)0.00391
1 x 1 x 0.5n \(\approx\)0.00391
You can solve this equation by using logarithms, the table or graph function on your calculator or by trial and improvement
n \(\approx \frac{\log0.00391}{\log0.5}\)
n = 8
The probability that Kylie scores a penalty in a game of rugby is 0.25. Find the least number of attempts that she would need to take to be 95% sure of scoring at least one penalty.
n =
X~B(n,0.25)
P(X\(\ge\)1) > 0.95
P(X = 0) < 0.05
nC0 x 0.250 x 0.75n < 0.05
0.75n < 0.05
Solve 0.75n = 0.05
n =\(\frac{\log 0.05}{\log0.75}\approx10.4\)
X~B(10,0.25) P(X\(\ge\)1) \(\approx\)0.944
X~B(11,0.25) P(X\(\ge\)1) \(\approx\)0.958
Therefore n = 11
The random variable X~B(n , p) has a mean = 5 and a variance = 4
Find n
n =
mean = np
variance = np(1-p)
np = 5
np(1-p) = 4
substitute the first equation into the second
5(1-p) = 4
1-p = \(\frac{4}{5}\)
p =\(\frac{1}{5}\)
A biased coin is flipped eight times. The probability of flipping a tail is 0.6.
What is the probability of flipping at least 6 tails given that we have flipped at least 4 tails?
Give your answer to 3 significant figures.
P =
X~B(8 , 0.6)
P(X \(\ge\) 6 \ X \(\ge\) 4)
\(\frac { P(X\ge 6) }{ P(X\ge 4) } \approx \frac { 0.3154 }{ 0.8263 } \approx 0.382\)
On average, it is found that 5% of AirPods* made on a production line are faulty.
a) Find the probability that in a random sample of 10, there are
i) No faulty AirPods
ii) more than one faulty set of AirPods
b) A sample of n sets of AirPods is taken from the production line. If the probability that there is at least one faulty AirPod is more than 75%, find the smallest possible value of n
* Fictitious data. I’m sure that AirPods production is very reliable!
Hint
Full Solution
Monster Energy drink cans are advertised to contain 500ml. During the production, the volume X in ml in cans of the drink can be modelled by a normal distribution with mean 504 and variance 10.
a) For a randomly selected can, work out P(X > 500).
b) Cans are put in packs of 6. Find the probability that at least 5 cans are have a volume of at least 500ml.
Hint
Full Solution
A glass contains 5 green sweets and m sweets of other colours. A sweet is taken at random.
a) Write down the probability that the sweet is green.
b) The sweet is replaced in the glass and the process is repeated a further three times. Each time, it is noted whether a green sweet is taken. The variance of the number of green sweets taken over the whole process in calculated to be 0.75.
Work out how many sweets in total there are in the glass.
Hint
Full Solution
How much of Binomial Distribution have you understood?
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