AI SL · 13.2
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The binomial distribution
13.2.4
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n trials, each a success with probability p. X is the number of successes. Choose which values of X to add up, or press a bar.
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Move a slider to change the distribution.
A bag holds 6 red counters and 4 green counters. Three counters are taken out at random, one after the other, and not put back. X is the number of red counters taken. Can X be modelled by a binomial distribution? First find the probability that the second counter is red, given that the first was red.
Have a go first. Then open the working one line at a time.
Yes: 3 counters.
A fixed number of trials is the first condition.
Yes: red or green.
Red is a success, green a failure.
P(1st red) = 6/10 = 0.6. P(2nd red | 1st red) = 5/9 = 0.556
After a red counter is removed, 5 reds are left among 9 counters. The probability changed.
Not binomial.
The trials are not independent and p is not constant, because the counters are not put back.
A large batch of light bulbs contains 8% that are faulty. A sample of 25 bulbs is chosen at random. X is the number of faulty bulbs in the sample. Show that X is binomial and write down its distribution. Give the value of p.
Have a go first. Then open the working one line at a time.
25 trials; faulty or not; independent (large batch); p = 0.08 each time.
A sample of 25 from a large batch changes the proportion by almost nothing, so the trials count as independent.
n = 25, p = 0.08
n is the number of trials. p is the probability of a success, the thing X counts.
X ~ B(25, 0.08)
Read it as: X is distributed binomially with n = 25 and p = 0.08.
Each loaf from a bakery is underbaked with probability 0.15, independently of the others. A box holds 12 loaves. X is the number of underbaked loaves in a box. Find the probability that exactly 3 loaves are underbaked.
Have a go first. Then open the working one line at a time.
X ~ B(12, 0.15)
State the distribution first.
From the calculator: binomial pdf with n = 12, p = 0.15, x = 3
Exactly one value of X: use binomial pdf.
P(X = 3) = 0.172
3 significant figures.
Use the same boxes of 12 loaves, X ~ B(12, 0.15). Find the probability that at most 2 loaves are underbaked.
Have a go first. Then open the working one line at a time.
From the calculator: binomial cdf with n = 12, p = 0.15, lower = 0, upper = 2
At most 2 adds the probabilities from 0 up to 2. Lower is 0, upper is 2.
P(X ≤ 2) = 0.736
The cdf adds up everything from the lower to the upper value.
Each seed in a packet germinates with probability 0.8, independently. A packet has 15 seeds. X is the number that germinate. Find the probability that at least 13 seeds germinate.
Have a go first. Then open the working one line at a time.
P(X ≥ 13) = 1 − P(X ≤ 12)
X is a whole number, so the values 13 and above are everything except 0 to 12.
From the calculator: binomial cdf with n = 15, p = 0.8, lower = 0, upper = 12
Lower is 0, upper is 12.
P(X ≤ 12) = 0.6020, so P(X ≥ 13) = 1 − 0.6020 = 0.398
A check: set lower = 13 and upper = 15 for the same answer.
A call centre makes 18 calls. Each call independently ends in a sale with probability 0.4. X is the number of sales. Find the probability that there are at least 5 but fewer than 9 sales.
Have a go first. Then open the working one line at a time.
P(5 ≤ X < 9) = P(5 ≤ X ≤ 8)
Fewer than 9 means up to 8.
From the calculator: binomial cdf with n = 18, p = 0.4, lower = 5, upper = 8
Set the lower value to 5 and the upper value to 8.
P(5 ≤ X ≤ 8) = 0.643
Check: P(X ≤ 8) − P(X ≤ 4) = 0.7368 − 0.09417.