The binomial distribution Problem set B One question at a time

For each situation, decide whether X can be modelled by a binomial distribution. If it can, write X ~ B(n, p). If it cannot, say which condition fails.

1. (a) A fair ten-sided die, numbered 1 to 10, is rolled 20 times. X is the number of rolls that show a multiple of 3.

   • 20 trials, two outcomes (multiple of 3 or not), independent, p = 3/10 = 0.3 each time
   • X ~ B(20, 0.3)

2. (b) A player takes free throws until she misses for the first time. X is the number of throws she takes.

   • The throws stop at the first miss, so there is no fixed number of trials

3. (c) A bag holds 10 tiles, 3 of them blue. Four tiles are taken out and not put back. X is the number of blue tiles taken.

   • The probability of blue changes after each tile is removed: 3/10, then 2/9 or 3/9
   • The trials are not independent

4. (d) In a large city, 18% of cars are silver. 40 cars pass a camera. X is the number of silver cars.

   • 40 trials, silver or not, independent (large city), p = 0.18 each time
   • X ~ B(40, 0.18)

\(X \sim B(19, 0.55)\). Find

1. (a) \(P(X = 11)\)

   • From the calculator: binomial pdf with n = 19, p = 0.55, x = 11
   • P(X = 11) = 0.177

2. (b) \(P(X \le 9)\)

   • From the calculator: binomial cdf with n = 19, p = 0.55, lower = 0, upper = 9
   • P(X ≤ 9) = 0.329

3. (c) \(P(X > 12)\)

   • P(X > 12) = 1 − P(X ≤ 12)
   • From the calculator: binomial cdf with n = 19, p = 0.55, lower = 0, upper = 12
   • P(X ≤ 12) = 0.8273
   • P(X > 12) = 1 − 0.8273 = 0.173

4. (d) \(P(6 < X \le 10)\)

   • P(6 < X ≤ 10) = P(7 ≤ X ≤ 10)
   • From the calculator: binomial cdf with n = 19, p = 0.55, lower = 7, upper = 10
   • P(7 ≤ X ≤ 10) = 0.472

\(X \sim B(40, 0.07)\). Find

1. (a) \(P(X < 5)\)

   • P(X < 5) = P(X ≤ 4)
   • From the calculator: binomial cdf with n = 40, p = 0.07, lower = 0, upper = 4
   • P(X ≤ 4) = 0.855

2. (b) \(P(X \ge 3)\)

   • P(X ≥ 3) = 1 − P(X ≤ 2)
   • From the calculator: binomial cdf with n = 40, p = 0.07, lower = 0, upper = 2
   • P(X ≤ 2) = 0.46252
   • P(X ≥ 3) = 1 − 0.46252 = 0.537

Each bus on a commuter's route is late with probability 0.11, independently of the other buses. She takes 18 buses in a month. X is the number of late buses.

1. (a) Write down the distribution of X.

   • 18 trials, late or not, independent, p = 0.11 each time
   • X ~ B(18, 0.11)

2. (b) Find the probability that exactly 3 buses are late.

   • From the calculator: binomial pdf with n = 18, p = 0.11, x = 3
   • P(X = 3) = 0.189

3. (c) Find the probability that at least 4 buses are late.

   • P(X ≥ 4) = 1 − P(X ≤ 3)
   • From the calculator: binomial cdf with n = 18, p = 0.11, lower = 0, upper = 3
   • P(X ≤ 3) = 0.8718
   • P(X ≥ 4) = 1 − 0.8718 = 0.128

4. (d) Find the probability that between 2 and 5 buses are late, inclusive.

   • From the calculator: binomial cdf with n = 18, p = 0.11, lower = 2, upper = 5
   • P(2 ≤ X ≤ 5) = 0.594

A class has 9 girls and 11 boys. A teacher chooses 5 pupils at random for a team, one after another, with no pupil chosen twice. X is the number of girls chosen.

1. (a) Explain why X is not binomial.

   • The probability of a girl changes after each pupil is removed
   • So p is not the same each time, and the trials are not independent

2. (b) Find the probability that the second pupil chosen is a girl, given that the first pupil was a girl. Compare it with the probability that the first pupil is a girl.

   • First pupil a girl: 9/20 = 0.450
   • Second a girl, given the first was a girl: 8/19 = 0.421
   • The probabilities differ, so the binomial conditions fail

Each person stopped in the street independently gives to a charity with probability 0.25. 30 people are stopped. X is the number who give.

1. (a) Find the probability that at least 4 but fewer than 9 people give.

   • P(4 ≤ X < 9) = P(4 ≤ X ≤ 8)
   • From the calculator: binomial cdf with n = 30, p = 0.25, lower = 4, upper = 8
   • P(4 ≤ X ≤ 8) = 0.636

2. (b) Find the probability that at least 10 people give.

   • P(X ≥ 10) = 1 − P(X ≤ 9)
   • From the calculator: binomial cdf with n = 30, p = 0.25, lower = 0, upper = 9
   • P(X ≤ 9) = 0.8034
   • P(X ≥ 10) = 1 − 0.8034 = 0.197

12% of the pupils in a school have flu. The flu spreads easily between pupils who share a room. 10 pupils from one room are chosen. X is the number with flu. Explain why X may not be binomial.

• Binomial trials must be independent
• Flu spreads between pupils who share a room, so one pupil having it raises the chance for the others
• The trials are not independent, so X may not be binomial