13.1 Discrete probability distributions and the binomial distribution — Practise

Work through these on your own. Write down $n$ , $p$ and $r$ before you substitute into the formula.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A fair coin is tossed 10 times. The random variable $X$ is the number of heads. Write down the distribution of $X$ and find $P(X = 5)$ .
The probabilities for a discrete random variable $X$ are $0.1$ , $0.3$ , $k$ and $0.2$ . Find $k$ .
A fair six-sided die is rolled once. The random variable $X$ is the score. Find $P(X \ge 5)$ .
The random variable $X \sim B(5, 0.3)$ . Find $P(X = 2)$ .
The random variable $X \sim B(6, \tfrac14)$ . Find $P(X = 2)$ .
The random variable $X \sim B(4, \tfrac13)$ . Find $P(X = 1)$ .
The random variable $X \sim B(5, \tfrac23)$ . Find $P(X = 4)$ .
A discrete random variable $X$ has probabilities $0.2$ , $0.3$ , $0.4$ and $0.1$ . Work out the sum of the four probabilities and state whether they form a valid distribution.
A bag holds 5 red counters and 3 blue counters. Two counters are taken without replacement. The random variable $X$ is the number of red counters. Explain why $X$ is not binomial, and find $P(X = 1)$ .
The random variable $X \sim B(8, 0.2)$ . Find $P(X = 0)$ .
The random variable $X \sim B(10, 0.9)$ . Find $P(X = 8)$ .
A multiple-choice test has 8 questions. Each question has 5 options, and exactly one option is correct. A student guesses every answer. The random variable $X$ is the number of correct answers. Write down the distribution of $X$ and find $P(X = 3)$ .

Answers and concise solutions

1. The distribution and the probability

Answer: $X \sim B(10, \tfrac12)$ , because there are 10 independent tosses and the probability of a head is $\tfrac12$ every time. Then $P(X = 5) = \dbinom{10}{5} \left(\tfrac12\right)^{10} = \tfrac{63}{256} \approx 0.2461$ .

2. The value of $k$

Answer: The probabilities cover every value of $X$ , so they add to 1. Then $k = 1 - (0.1 + 0.3 + 0.2) = 1 - 0.6 = 0.4$ .

3. The probability

Answer: The scores 5 and 6 are the two outcomes that satisfy $X \ge 5$ , so $P(X \ge 5) = \tfrac{2}{6} = \tfrac13$ . The distribution has four other values, 1, 2, 3 and 4, and the six probabilities add to 1.

4. The probability

Answer: $P(X = 2) = \dbinom{5}{2}(0.3)^2(0.7)^3 = 10 \times 0.09 \times 0.343 = 0.3087$ . The two powers add to 5, which is $n$ .

5. The probability

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

6. The probability

Answer: $P(X = 1) = \dbinom{4}{1} \left(\tfrac13\right) \left(\tfrac23\right)^3 = 4 \times \tfrac13 \times \tfrac{8}{27} = \tfrac{32}{81} \approx 0.3951$ . The failure power is $4 - 1 = 3$ .

7. The probability

Answer: $P(X = 4) = \dbinom{5}{4} \left(\tfrac23\right)^4 \left(\tfrac13\right) = 5 \times \tfrac{16}{81} \times \tfrac13 = \tfrac{80}{243} \approx 0.3292$ .

8. The sum

Answer: The sum is $0.2 + 0.3 + 0.4 + 0.1 = 1$ . Every probability lies between 0 and 1, and the sum is 1, so these four probabilities form a valid probability distribution.

9. The probability

Answer: The probability of red changes from $\tfrac58$ on the first draw to $\tfrac47$ or $\tfrac57$ on the second, so $p$ is not constant and the binomial model does not apply. The two orders give $P(X = 1) = \tfrac58 \times \tfrac37 + \tfrac38 \times \tfrac57 = \tfrac{15}{28} \approx 0.5357$ .

10. The probability

Answer: $P(X = 0) = \dbinom{8}{0}(0.2)^0(0.8)^8 = (0.8)^8 = \tfrac{65536}{390625} \approx 0.1678$ . Every trial is a failure.

11. The probability

Answer: $P(X = 8) = \dbinom{10}{8}(0.9)^8(0.1)^2 = 45 \times 0.43046721 \times 0.01 \approx 0.1937$ . The failure power is $10 - 8 = 2$ .

12. The distribution and the probability

Answer: $X \sim B(8, \tfrac15)$ , because there are 8 independent guesses and the probability of a correct guess is $\tfrac15$ each time. Then $P(X = 3) = \dbinom{8}{3} \left(\tfrac15\right)^3 \left(\tfrac45\right)^5 = \tfrac{57344}{390625} \approx 0.1468$ .