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Probability & distributions · Lesson 2 of 3

The binomial distribution

Repeated independent trials with the same chance of success: P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, choosing p and q correctly, and 'at least', 'at most' and 'more than' with the complement.

18 minYou should already know: Probability Permutation & combination
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Many questions repeat the same trial: 6 drivers checked, 10 questions guessed, 8 families surveyed. When each trial is independent and has the same chance of “success”, the number of successes follows a binomial distribution.

The formula

Let nn be the number of trials, pp the probability of success in one trial and q=1−pq = 1 - p. The probability of exactly rr successes is

P(X=r)=nCr prqn−rP(X = r) = {}^nC_r\,p^r q^{n - r}

Each arrangement of rr successes and n−rn - r failures has probability prqn−rp^r q^{n - r}, and there are nCr{}^nC_r ways to choose which trials succeed (see permutations and combinations).

012345X ~ B(5, 0.4)P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ
A binomial distributionOne bar for each number of successes

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q13

A survey conducted revealed that four out of every twenty taxi drivers do not have a valid driving licence. If 6 drivers are selected at random, calculate, correct to three decimal places, the probability that:

exactly 2;

more than 3;

at least 5 have valid driving licence.

  1. What is a success?

    • 4 in 20 have no valid licence, so q=420=0.2{q = \frac{4}{20} = 0.2}.
    • A success is a valid licence: p=0.8{p = 0.8}, with n=6{n = 6}.

    Think first. The question counts drivers with a valid licence. What is p?

  2. Exactly 2

    • P(X=2)=15×0.82×0.24{P(X = 2) = 15 \times 0.8^2 \times 0.2^4}.
    • =15×0.64×0.0016=0.01536≈0.015{= 15 \times 0.64 \times 0.0016 = 0.01536 \approx 0.015}.

    Think first. ⁶C₂ p² q⁴.

  3. More than 3

    • More than 3 means X=4,5{X = 4, 5} or 6{6}.
    • P(4)=15×0.84×0.22=0.24576{P(4) = 15 \times 0.8^4 \times 0.2^2 = 0.24576}.
    • P(5)=6×0.85×0.2=0.393216{P(5) = 6 \times 0.8^5 \times 0.2 = 0.393216} and P(6)=0.86=0.262144{P(6) = 0.8^6 = 0.262144}.
    • Add: 0.90112≈0.901{0.90112 \approx 0.901}.

    Think first. Which values of X?

  4. At least 5

    • P(5)+P(6)=0.393216+0.262144{P(5) + P(6) = 0.393216 + 0.262144}.
    • =0.65536≈0.655{= 0.65536 \approx 0.655}.

At least, at most, more than

Turn the words into values of XX first:

WordsValues of XX (out of nn)
at least 33,4,…,n3, 4, \ldots, n
at most 30,1,2,30, 1, 2, 3
more than 34,5,…,n4, 5, \ldots, n
fewer than 30,1,20, 1, 2

When the list is long, use the complement: “at least 3” is 1−[P(0)+P(1)+P(2)]1 - [P(0) + P(1) + P(2)].

The binomial distributionSet n and p, pick a condition
012345678
0.7447P(X ≥ 2)0.74471 − P(X ≤ 1)
Each bar is P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ with q = 0.7. "At least 2" is 7 bars; the 2 bars below it add to 0.2553, so it is quicker as 1 − 0.2553.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q12

In an examination, 60%60\% of the candidates passed. If 10 candidates are selected at random, find, correct to four decimal places, the probability that:

at least two of them failed;

exactly half of them passed;

at most two of them failed.

  1. Count failures

    • Let FF be the number who failed: F∼B(10,0.4){F \sim B(10, 0.4)}.
    • P(F=0)=0.610=0.0060{P(F = 0) = 0.6^{10} = 0.0060} and P(F=1)=10×0.4×0.69=0.0403{P(F = 1) = 10 \times 0.4 \times 0.6^9 = 0.0403}.

    Think first. Parts (a) and (c) are about failures. What is p for failing?

  2. At least two failed

    • P(F=0)+P(F=1)=0.0464{P(F = 0) + P(F = 1) = 0.0464}.
    • So P(F≥2)=1−0.0464=0.9536{P(F \ge 2) = 1 - 0.0464 = 0.9536}.

    Think first. Use the complement.

  3. Exactly half passed

    • 10C5×0.65×0.45{{}^{10}C_5 \times 0.6^5 \times 0.4^5}.
    • =252×0.07776×0.01024{= 252 \times 0.07776 \times 0.01024}.
    • ≈0.2007{\approx 0.2007}.

    Think first. 5 passed means 5 failed.

  4. At most two failed

    • P(F=2)=45×0.42×0.68=0.1209{P(F = 2) = 45 \times 0.4^2 \times 0.6^8 = 0.1209}.
    • P(F≤2)=0.0060+0.0403+0.1209{P(F \le 2) = 0.0060 + 0.0403 + 0.1209}.
    • =0.1673{= 0.1673}, to four decimal places.

More: the binomial distribution

Your turn

WAEC 2022 · Paper 2 · Q13

A soldier fires at a target and the probability of hitting the target with any shot is 25\frac25. If he fires 6 shots, find, correct to three decimal places, the probability that he hits the target:

  1. (a)

    6 times;

  2. (b)

    at most 3 times;

  3. (c)

    at least 2 times.

Worked solution (try it first)
  1. X∼B(6,25)X \sim B\left(6, \frac25\right).
  2. Over 15 625: P(0)=729P(0) = 729, P(1)=2916P(1) = 2916, P(2)=4860P(2) = 4860, P(3)=4320P(3) = 4320.

(a)

  1. P(6)=(25)6P(6) = \left(\frac25\right)^6
    =6415625= \dfrac{64}{15625}
    ≈0.004\approx 0.004.

(b)

  1. At most 3: 729+2916+4860+432015625=1282515625\dfrac{729 + 2916 + 4860 + 4320}{15625} = \dfrac{12825}{15625}
    ≈0.821\approx 0.821.

(c)

  1. At least 2: 1−729+291615625=11980156251 - \dfrac{729 + 2916}{15625} = \dfrac{11980}{15625}
    ≈0.767\approx 0.767.

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