13.1 Discrete probability distributions and the binomial distribution — Practise with help

Work each question closed book first. State what a success means before you use the binomial formula.

Warm-up: remember

The events $A$ and $B$ are independent, with $P(A) = 0.3$ and $P(B) = 0.5$ . Find $P(A \text{ and } B)$ .
The probability that it rains on a day is $0.42$ . Find the probability that it does not rain.
Work out $\dbinom{5}{2}$ .

Answers

For independent events, multiply: $P(A \text{ and } B) = 0.3 \times 0.5 = 0.15$ .
Use the complement: $1 - 0.42 = 0.58$ .
$\dbinom{5}{2} = \dfrac{5!}{2!\,3!} = \dfrac{5 \times 4}{2 \times 1} = 10$ .

If question 3 was hard, revise the binomial coefficient from Section 4.1.

Practise with help

1. A fair coin is tossed 8 times. The random variable $X$ is the number of heads. Write down the distribution of $X$ in the form $X \sim B(n, p)$ .

Hint

Check the four conditions. Then decide what a success is, and count the trials.

Next step

There are 8 tosses, so $n = 8$ . A head is a success, and the coin is fair, so $p = \tfrac12$ .

Full working

Each toss has two outcomes, the probability of a head is the same for every toss, and the tosses are independent. There are 8 tosses, so

X \sim B\left(8, \tfrac12\right).

2. The probabilities for a discrete random variable $X$ are $0.15$ , $0.25$ , $0.35$ and $k$ . Find $k$ .

Hint

The four probabilities cover every possible value, so they add to 1.

Next step

Add the three known probabilities first: $0.15 + 0.25 + 0.35 = 0.75$ .

Full working

The probabilities add to 1, so

k = 1 - (0.15 + 0.25 + 0.35) = 1 - 0.75 = 0.25.

3. The random variable $X \sim B(5, 0.3)$ . Find $P(X = 3)$ .

Hint

Write down $n$ , $r$ , $p$ and $1 - p$ before you substitute.

Next step

Here $n = 5$ , $r = 3$ , $p = 0.3$ and $1-p = 0.7$ . Substitute into $P(X = r) = \dbinom{n}{r} p^r (1-p)^{n-r}$ .

Full working

P(X = 3) = \binom{5}{3} (0.3)^3 (0.7)^{2}.

Work out each factor:

\binom{5}{3} = 10, \qquad (0.3)^3 = 0.027, \qquad (0.7)^{2} = 0.49.

Multiply:

P(X = 3) = 10 \times 0.027 \times 0.49 = 0.1323.

4. A bag holds 4 red counters and 2 blue counters. Two counters are taken without replacement. The random variable $X$ is the number of red counters. Explain why $X$ is not a binomial variable.

Hint

Look at the probability of a red counter on the first draw and on the second draw.

Next step

After one red counter is removed, there are 3 red and 2 blue left, so the second probability is different from the first.

Full working

The probability of red on the first draw is $\tfrac46 = \tfrac23$ . If the first counter is red, the probability of red on the second draw is $\tfrac35$ . The probability changes between the trials, so $p$ is not constant. The four binomial conditions are not all met, so $X$ is not a binomial variable. Taking counters without replacement from a small bag makes the trials dependent.

5. The random variable $X \sim B(4, \tfrac13)$ . Find $P(X = 1)$ .

Hint

Use $p = \tfrac13$ and $1 - p = \tfrac23$ . Keep the answer as a fraction.

Next step

Substitute $n = 4$ , $r = 1$ : $P(X = 1) = \dbinom{4}{1} \left(\tfrac13\right)^1 \left(\tfrac23\right)^3$ . Work out $\dbinom{4}{1}$ and $\left(\tfrac23\right)^3$ .

Full working

P(X = 1) = \binom{4}{1} \left(\tfrac13\right) \left(\tfrac23\right)^3 = 4 \times \tfrac13 \times \tfrac{8}{27}.

Multiply the fractions:

P(X = 1) = \tfrac{32}{81} \approx 0.3951.

6. The random variable $X \sim B(6, \tfrac14)$ . Find $P(X = 2)$ , and explain the part played by the number 15 in your working.

Hint

The number 15 is $\dbinom{6}{2}$ . Think about what it counts.

Next step

Substitute $n = 6$ , $r = 2$ , $p = \tfrac14$ and $1-p = \tfrac34$ : $P(X = 2) = \dbinom{6}{2} \left(\tfrac14\right)^2 \left(\tfrac34\right)^4$ . Work out the coefficient and the two powers.

Full working

P(X = 2) = \binom{6}{2} \left(\tfrac14\right)^2 \left(\tfrac34\right)^4 = 15 \times \tfrac{1}{16} \times \tfrac{81}{256} = \tfrac{1215}{4096} \approx 0.2966.

The number 15 counts the positions of the two successes among the six trials. There are 15 different pairs of trials that could hold the two successes, and each arrangement has the same probability.