To answer questions involving the binomial probability distribution, we need the technical information about the distribution that was presented in an earlier chapter.
Situations that may be modelled by a binomial distribution are those in which there are a number of independent trials of the same experiment and the individual outcomes are either success or failure. We are interested in the probability that there will be some number $r$r successes among the $n$n trials of the experiment.
For example, the 'experiment' might be a survey in which the respondents can answer either yes or no. Or, it might be it might be a chicken-hatching experiment with a batch of $n$n eggs in which individuals eventually either hatch or fail to hatch.
Example
Continuing with the chicken-hatching example, we might know from previous observations that the probability of an egg hatching is $p$p. We would expect $np$np chickens to hatch and would not be surprised to see some variation around this mean value. The variance, $np\left(1-p\right)$np(1−p) and the standard deviation, which is the square root of the variance, gives an indication of how much variability to expect.
If a particular batch of eggs had a hatch rate that was more than, say, one standard deviation away from the mean, we might wonder whether something unusual had happened with the process.
Suppose the usual hatch rate is $60%$60% and a batch of $40$40 eggs is being incubated. The mean or expected number of chicks would be $40\times0.6=24$40×0.6=24.
The standard deviation in this case is $\sqrt{40\times0.6\times0.4}\approx3.1$√40×0.6×0.4≈3.1. We might be alarmed if fewer than $21$21 of the eggs hatched.
The probability of hatching exactly $20$20 chicks can be found using the binomial probability formula
$P(N=r)=\binom{n}{r}p^r\left(1-p\right)^{n-r}$P(N=r)=(nr)pr(1−p)n−r
In this case, we have $P(N=20)=\binom{40}{20}\times0.6^{20}\times\left(1-0.6\right)^{40-20}$P(N=20)=(4020)×0.620×(1−0.6)40−20 and this simplifies to the probability $0.055$0.055. We could calculate in the same way all the probabilities from zero up to $20$20 and add these up to find the probability of obtaining fewer than $21$21 chicks. (This is quite tedious and is best done by machine. Use of a spreadsheet application is one solution.)
In this way, we find that the probability of $20$20 or fewer chicks hatching is $0.13$0.13 while the probability of between $21$21 and $27$27 hatching (that is, a number within a one-standard-deviation range about the mean) is $0.74$0.74.
Worked Examples
QUESTION 1
A particular nationwide numeracy test has a failure rate of $30%$30%.
If you randomly selected $100$100 students from across the country to do the test, how many would you expect to pass?
QUESTION 2
Tom is currently applying for graduate job positions. For each application he submits, the probability that it gets short-listed and he gets invited for an interview is $0.01$0.01.
If he applies for $8$8 positions, what is the probability that he will not get a single interview?
Give your answer correct to two decimal places.
If he applies for $8$8 positions, what is the probability that he will get at least one interview?
Give your answer correct to two decimal places.
If he applies for $n$n positions, what is the probability that he will get at least one interview?
Hence find the minimum number of applications he will have to submit to ensure that the probability that he gets at least one interview is greater than $0.9$0.9.
QUESTION 3
Records show that half of all households in a city have broadband. What is the probability that less than $45%$45% of a sample of $100$100 random households have a broadband?
Give your answer as a decimal correct to four decimal places.