AS & A Level · Practice questions · The binomial distribution

Fixed trials, fixed p.

Six original Cambridge-style questions on the binomial distribution, at Paper 5 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Define \(X\), then \(\text{B}(n, p)\).

Model: state \(X\), \(n\) and \(p\), and check the four conditions.
Exact: \(\text{P}(X = r) = \dbinom{n}{r} p^{r}(1-p)^{n-r}\).
Cumulative: turn the words into an inequality; use the complement when shorter.
Mean and variance: \(np\) and \(np(1-p)\).
Least n: set up the inequality and try values either side.

Recall

[2 marks]

A fair die is thrown 15 times and \(X\) is the number of sixes. State the distribution of \(X\) and give two conditions needed for it to be valid.

\(X \sim \text{B}(15, \tfrac{1}{6})\). ✓
The throws are independent, and the probability of a six is the same (\(\tfrac{1}{6}\)) on every throw. ✓

Calculation

[4 marks]

\(X \sim \text{B}(15, 0.2)\). Find (a) \(\text{P}(X = 4)\), (b) \(\text{P}(X < 3)\).

(a) \(\dbinom{15}{4}(0.2)^4(0.8)^{11} = 1365 \times 0.0016 \times 0.085\,90 = 0.188\) (3 s.f.). ✓
(b) \(\text{P}(0) + \text{P}(1) + \text{P}(2) = 0.035\,18 + 0.131\,94 + 0.230\,89 = 0.398\) (3 s.f.). ✓

Calculation

[4 marks]

\(X \sim \text{B}(n, p)\) has mean 4.8 and variance 2.88. Find \(n\) and \(p\), and hence find \(\text{P}(X = 5)\).

\(1 - p = 2.88 \div 4.8 = 0.6\), so \(p = 0.4\) and \(n = 4.8 \div 0.4 = 12\). ✓
\(\text{P}(X = 5) = \dbinom{12}{5}(0.4)^5(0.6)^7 = 792 \times 0.010\,24 \times 0.027\,99 = 0.227\) (3 s.f.). ✓

Calculation

[3 marks]

A fair die is thrown \(n\) times. Find the least value of \(n\) for which the probability of at least one six is greater than 0.9.

Need \(1 - \left(\tfrac{5}{6}\right)^n > 0.9\), i.e. \(\left(\tfrac{5}{6}\right)^n < 0.1\). ✓
\(\left(\tfrac{5}{6}\right)^{12} = 0.112\): too big. \(\left(\tfrac{5}{6}\right)^{13} = 0.0935 < 0.1\). ✓
Least \(n = 13\). ✓

Analysis

[5 marks]

A multiple-choice test has 20 questions, each with 4 options, only one of which is correct. A student guesses every answer.

(a) Find the expected number of correct answers and the variance. (b) The pass mark is 8. Find the probability that the student passes.

(a) \(X \sim \text{B}(20, 0.25)\): \(\text{E}(X) = 5\), \(\text{Var}(X) = 20 \times 0.25 \times 0.75 = 3.75\). ✓
(b) \(\text{P}(X \ge 8) = 1 - \text{P}(X \le 7)\). ✓
\(\text{P}(X \le 7) = 0.8982\) (adding the eight terms from 0 to 7). ✓
\(\text{P}(X \ge 8) = 0.102\) (3 s.f.). ✓

Analysis

[6 marks]

A light bulb passes a quality check with probability 0.9, independently of other bulbs. Bulbs are packed in boxes of 10, and a box is 'good' if at least 9 of its bulbs pass.

(a) Find the probability that a box is good. (b) A shop orders 6 boxes. Find the probability that at least 5 of them are good.

(a) \(X \sim \text{B}(10, 0.9)\): \(\text{P}(9) + \text{P}(10) = 10(0.9)^9(0.1) + 0.9^{10} = 0.3874 + 0.3487 = 0.7361\), so 0.736 (3 s.f.). ✓
(b) \(Y \sim \text{B}(6, 0.7361)\). ✓
\(\text{P}(Y = 5) = 6(0.7361)^5(0.2639) = 0.3422\); \(\text{P}(Y = 6) = (0.7361)^6 = 0.1591\). ✓
\(\text{P}(Y \ge 5) = 0.501\) (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.