Mathematics 9709/61 — October/November 2015

Cambridge AS Level · Probability & Statistics 1 · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Discrete Random Variables · The Normal Distribution · Representation of Data · Permutations and Combinations · Probability

Q14MDiscrete Random VariablesFree sample

In a certain town, 76% of cars are fitted with satellite navigation equipment. A random sample of 11 cars from this town is chosen. Find the probability that fewer than 10 of these cars are fitted with this equipment.

DifficultyMedium-Easy

Worked solution

Approach

Let $X$ be the number of cars in the sample fitted with satellite navigation. Each car independently has probability $p = 0.76$ of being fitted, so $X \sim B(11, 0.76)$ . We require $P(X < 10)$ , i.e. the probability that $X$ is 9 or fewer. Since this event has many terms, use the complement: the only outcomes that are not "fewer than 10" are $X = 10$ and $X = 11$ , so $P(X < 10) = 1 - P(X = 10) - P(X = 11)$ .

Working

The probability of exactly $x$ successes is given by the binomial formula

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}

Here $n = 11$ and $p = 0.76$ .

"Fewer than 10" means $X = 0, 1, \ldots, 9$ , so

P(X < 10) = 1 - P(X \ge 10) = 1 - P(X = 10) - P(X = 11)

For $X = 10$ :

P(X = 10) = \binom{11}{10}(0.76)^{10}(0.24)^{1} = 11(0.76)^{10}(0.24)

For $X = 11$ :

P(X = 11) = \binom{11}{11}(0.76)^{11}(0.24)^{0} = (0.76)^{11}

Therefore

P(X < 10) = 1 - 11(0.76)^{10}(0.24) - (0.76)^{11} = 1 - 0.219

P(X < 10) = 0.781

Answer

0.781

Final answer

0.781

Detailed explanation

Walkthrough

This is a binomial distribution problem. We have a fixed number of independent trials, $n = 11$ (the sample size), and each trial has the same success probability $p = 0.76$ (the probability a car has satellite navigation). Define $X$ as the number of cars in the sample that are fitted, so $X \sim B(11, 0.76)$ .

We want the probability that fewer than 10 cars are fitted, i.e. $P(X < 10) = P(X = 0) + P(X = 1) + \cdots + P(X = 9)$ . Adding ten separate terms is slow and error-prone, so instead we use the complement rule. The event "fewer than 10" fails precisely when $X = 10$ or $X = 11$ , so

P(X < 10) = 1 - P(X = 10) - P(X = 11)

For $X = 10$ , the binomial formula gives

P(X = 10) = \binom{11}{10}(0.76)^{10}(0.24)^{1} = 11(0.76)^{10}(0.24)

because there are 10 successes and 1 failure. For $X = 11$ , all 11 cars are fitted, so

P(X = 11) = \binom{11}{11}(0.76)^{11}(0.24)^{0} = (0.76)^{11}

Adding these two together and subtracting from 1 gives $1 - 0.219 = 0.781$ . The answer is a probability, so it lies between 0 and 1 and is reported to three significant figures.

Key Takeaways

Recognise a binomial distribution when there is a fixed number of independent trials and a fixed success probability.
The complement rule $P(X < k) = 1 - P(X \ge k)$ drastically reduces the amount of working when the complement has only a few outcomes.
Use the binomial formula $P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$ for each individual outcome.

Common Mistakes

Confusing "fewer than 10" with "at most 10": "fewer than 10" excludes $X = 10$ and $X = 11$ , whereas "at most 10" excludes only $X = 11$ .
Trying to add all ten terms $P(X = 0)$ up to $P(X = 9)$ directly, which is wasteful and invites arithmetic slips.
Getting the failure exponent wrong: for $X = 10$ with $n = 11$ , the failure term is $(0.24)^1$ , not $(0.24)^0$ .
Forgetting that for $X = 11$ the term is simply $(0.76)^{11}$ with no factor of $0.24$ .

Things to Be Careful About

The complement of $X < 10$ is $X \ge 10$ , i.e. $X = 10$ or $X = 11$ — not $X > 10$ .
Check that $0.76 + 0.24 = 1$ ; the failure probability is $1 - p = 1 - 0.76 = 0.24$ .
The final probability is $1 - 0.219 = 0.781$ . Give it to three significant figures as in the mark scheme, and make sure the value lies between 0 and 1.

Techniques used

model independent trials with a binomial distributionapply the complement rule to a cumulative probabilitycompute individual binomial probability termsevaluate probabilities to obtain a final decimal answer

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