13.2 Calculating binomial probabilities — Practise

Work through these on your own. Write the inequality in words first, then choose the shorter route. Round once, at the end.

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

A biased coin lands on heads with probability 0.4. It is tossed 10 times. The random variable $X$ is the number of heads. Find $P(X \le 2)$ .
A basketball player scores a free throw with probability 0.4. She takes 10 free throws. The random variable $X$ is the number she scores. Find $P(X \ge 8)$ .
A machine produces faulty items with probability 0.3. A sample of 12 items is checked. The random variable $X$ is the number of faulty items. Find $P(3 \le X \le 5)$ .
A fair coin is tossed 20 times. The random variable $X$ is the number of heads. Find $P(X \le 9)$ .
A quiz has 15 questions. Each question has 5 options, and exactly one is correct. A student guesses every answer. The random variable $X$ is the number correct. Find $P(X \ge 4)$ .
A seed germinates with probability 0.6. Eight seeds are planted. The random variable $X$ is the number that germinate. Find $P(X \ge 1)$ .
Each of 20 seeds germinates with probability 0.8, independently. The random variable $X$ is the number that germinate. Find $P(X \le 13)$ .
A fair coin is tossed 6 times. The random variable $X$ is the number of heads. Find $P(X \ge 5)$ .
A machine produces faulty items with probability 0.1. A sample of 50 items is checked. The random variable $X$ is the number of faulty items. Find $P(X = 0)$ .
A spinner lands on red with probability $\tfrac13$ . It is spun 9 times. The random variable $X$ is the number of reds. Find $P(X \le 1)$ .
A fair coin is tossed 20 times. The random variable $X$ is the number of heads. Find $P(8 \le X \le 12)$ .
A test has 20 questions. Each has 4 options, and exactly one is correct. A student guesses every answer. The random variable $X$ is the number correct. Find $P(X \ge 10)$ .

Answers and concise solutions

1. The probability

Answer: $P(X \le 2) = P(0) + P(1) + P(2) = 0.0060 + 0.0403 + 0.1209 = 0.1673$ (4 d.p.). Adding the three terms is shorter than using the complement.

2. The probability

Answer: Adding the terms for 8, 9 and 10 is shorter than the complement: $0.0106 + 0.0016 + 0.0001 = 0.0123$ (4 d.p.).

3. The probability

Answer: Add the terms for 3, 4 and 5: $0.2397 + 0.2311 + 0.1585 = 0.6293$ (4 d.p.). Both ends are included.

4. The probability

Answer: Every term has denominator $2^{20} = 1048576$ . The terms for 0 to 9 add to $\tfrac{431910}{1048576} = \tfrac{215955}{524288} \approx 0.4119$ .

5. The probability

Answer: $X \sim B(15, \tfrac15)$ . Use the complement: $P(X \ge 4) = 1 - P(X \le 3) = 1 - 0.6482 = 0.3518$ (4 d.p.).

6. The probability

Answer: The complement of "at least one" is "none". So $P(X \ge 1) = 1 - (0.4)^8 = 1 - \tfrac{65536}{390625} = \tfrac{390369}{390625} \approx 0.9993$ .

7. The probability

Answer: $P(X \le 13) = 1 - P(X \ge 14) = 1 - (P(14) + \dots + P(20)) \approx 0.0867$ . So 13 or fewer seeds germinate in about 8.7% of packets of this kind.

8. The probability

Answer: $P(X \ge 5) = P(5) + P(6) = \tfrac{6}{64} + \tfrac{1}{64} = \tfrac{7}{64} \approx 0.1094$ . There are only two terms, so the direct sum is shorter than the complement.

9. The probability

Answer: $P(X = 0) = (0.9)^{50} \approx 0.0052$ . Every item must pass, and each one does so with probability 0.9.

10. The probability

Answer: $P(X \le 1) = P(0) + P(1) = \left(\tfrac23\right)^9 + 9\left(\tfrac13\right)\left(\tfrac23\right)^8 \approx 0.1431$ . Keep the terms to four decimal places and round once.

11. The probability

Answer: Add the five terms for 8, 9, 10, 11 and 12: $P(8 \le X \le 12) = \tfrac{96577}{131072} \approx 0.7368$ . The distribution is symmetric about 10, so the terms for 8 and 12 are equal, and so are those for 9 and 11.

12. The probability

Answer: $X \sim B(20, \tfrac14)$ . Use the complement: $P(X \ge 10) = 1 - P(X \le 9) = 1 - 0.9861 = 0.0139$ (4 d.p.). Guessing as many as 10 is unlikely.