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The binomial distributionEdexcel A-Level Maths: Revision notes

Section 1

The binomial model

A binomial distribution X∼B(n,p)X\sim B(n,p) models the number of successes in nn trials when: there is a fixed number of trials nn; each trial has two outcomes (success or failure); the probability of success pp is constant; the trials are independent. Examples: the number of correct guesses in a 10-question test with four options (p=0.25p=0.25), or defective bolts in a sample of 20 (p=0.05p=0.05). State the distribution with its parameters, and justify the model by naming these conditions.

Key termsbinomial distributiontrialsuccess
Common mistake

Using a binomial model when the probability changes, for example drawing without replacement from a small bag.

Section 2

Individual probabilities

For X∼B(n,p)X\sim B(n,p), P(X=r)=(nr)pr(1−p)n−r,r=0,1,…,n.P(X=r)=\binom{n}{r}p^r(1-p)^{n-r},\qquad r=0,1,\dots,n. (nr)\binom nr counts the ways of choosing which rr trials are successes. Example: X∼B(10,0.25)X\sim B(10,0.25) gives P(X=3)=(103)(0.25)3(0.75)7=0.250P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7=0.250. Use the calculator's binomial probability density function for P(X=r)P(X=r), and check you have not left out the (nr)\binom nr factor.

Key termsprobability formula

Section 3

Cumulative probabilities

The calculator's cumulative function gives P(X≤r)P(X\le r). Convert other inequalities to this form, using the fact that XX takes whole numbers only:

  • P(X<r)=P(X≤r−1)P(X<r)=P(X\le r-1)
  • P(X≥r)=1−P(X≤r−1)P(X\ge r)=1-P(X\le r-1)
  • P(X>r)=1−P(X≤r)P(X>r)=1-P(X\le r)
  • P(a≤X≤b)=P(X≤b)−P(X≤a−1)P(a\le X\le b)=P(X\le b)-P(X\le a-1). Example: X∼B(10,0.25)X\sim B(10,0.25), P(X≥4)=1−P(X≤3)=1−0.7759=0.224P(X\ge4)=1-P(X\le3)=1-0.7759=0.224. 'At least one' is quickest as 1−P(X=0)=1−(1−p)n1-P(X=0)=1-(1-p)^n.
Key termscumulative probability
Common mistake

Writing P(X≥4)=1−P(X≤4)P(X\ge4)=1-P(X\le4). That removes X=4X=4 as well; it should be 1−P(X≤3)1-P(X\le3).

Section 4

Two-stage and inequality problems

Sometimes a binomial probability becomes the pp of a second binomial. Example: a sample is accepted if X≤2X\le2 with q=P(X≤2)=0.9245q=P(X\le2)=0.9245. The number accepted from 5 samples is W∼B(5,0.9245)W\sim B(5,0.9245), so P(W=4)=5q4(1−q)=0.276P(W=4)=5q^4(1-q)=0.276. To find the least nn for a target, form an inequality: 1−0.4n>0.991-0.4^n>0.99 gives 0.4n<0.010.4^n<0.01, so nlog⁡0.4<log⁡0.01n\log0.4<\log0.01 and, dividing by the negative log⁡0.4\log0.4, n>5.03n>5.03, so n=6n=6. Also find the largest kk with P(X≥k)>0.9P(X\ge k)>0.9 by evaluating cumulative probabilities for neighbouring values.

Key termsinequality
Exam tip

Check an inequality answer by testing nn and n−1n-1: 0.45=0.01020.4^5=0.0102 is too big, 0.46=0.00410.4^6=0.0041 works.

Section 5

Validity of the model

Binomial answers are only reliable if the conditions hold. Independence may fail when trials share a cause, such as students in one class or consecutive throws affected by fatigue. The probability may vary between trials. Comment in context: 'independence is unlikely because ...' and say what difference it would make.

Exam tip

In a 'state two assumptions' question, write 'trials are independent' and 'probability of success is constant', each in context.

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Exam questions on The binomial distribution

  1. A student guesses the answer to every question on a 10-question multiple-choice test. Each question has four options, exactly one of which is correct. The random variable XX is the number of questions answered correctly, and X∼B(10,0.25)X\sim B(10,0.25).
    Find P(X≥4)P(X\ge4).2 marks
  2. A gardener plants 12 seeds. Each seed germinates with probability 0.90.9, independently of the others. The random variable YY is the number of seeds that germinate.
    Find P(Y=10)P(Y=10).2 marks
  3. A machine produces bolts. Each bolt is defective with probability 0.050.05, independently of the others. Bolts are packed in samples of 20, and XX is the number of defective bolts in a sample.
    State the distribution of XX and find the probability that a sample contains exactly 2 defective bolts.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).