The binomial distribution Problem set A Worked solutions

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 eight-sided die is rolled 24 times. X is the number of 5s.

   • 24 trials, two outcomes (5 or not 5), independent, p = 1/8 = 0.125 each time
   • X ~ B(24, 0.125)

2. (b) Cards are drawn one at a time from a pack, each card being put back and the pack shuffled. X is the number of cards drawn until the first heart appears.

   • The drawing stops at the first heart, so there is no fixed number of trials

3. (c) A box holds 9 chocolates, 4 of them dark. Three chocolates are chosen at random and eaten. X is the number of dark chocolates chosen.

   • The probability of dark changes after each chocolate is removed: 4/9, then 3/8 or 4/8
   • The trials are not independent

4. (d) In a large production run, 5% of tablets are faulty. 30 tablets are picked at random. X is the number that are faulty.

   • 30 trials, faulty or not, independent (large production run), p = 0.05 each time
   • X ~ B(30, 0.05)

\(X \sim B(16, 0.4)\). Find

1. (a) \(P(X = 6)\)

   • From the calculator: binomial pdf with n = 16, p = 0.4, x = 6
   • P(X = 6) = 0.198

2. (b) \(P(X \le 5)\)

   • From the calculator: binomial cdf with n = 16, p = 0.4, lower = 0, upper = 5
   • P(X ≤ 5) = 0.329

3. (c) \(P(X > 9)\)

   • P(X > 9) = 1 − P(X ≤ 9)
   • From the calculator: binomial cdf with n = 16, p = 0.4, lower = 0, upper = 9
   • P(X ≤ 9) = 0.9417
   • P(X > 9) = 1 − 0.9417 = 0.0583

4. (d) \(P(3 < X \le 7)\)

   • P(3 < X ≤ 7) = P(4 ≤ X ≤ 7)
   • From the calculator: binomial cdf with n = 16, p = 0.4, lower = 4, upper = 7
   • P(4 ≤ X ≤ 7) = 0.651

\(X \sim B(30, 0.12)\). Find

1. (a) \(P(X < 4)\)

   • P(X < 4) = P(X ≤ 3)
   • From the calculator: binomial cdf with n = 30, p = 0.12, lower = 0, upper = 3
   • P(X ≤ 3) = 0.507

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

   • P(X ≥ 2) = 1 − P(X ≤ 1)
   • From the calculator: binomial cdf with n = 30, p = 0.12, lower = 0, upper = 1
   • P(X ≤ 1) = 0.1100
   • P(X ≥ 2) = 1 − 0.1100 = 0.890

In a large delivery of mugs, 7% are chipped. A restaurant picks 20 mugs at random. X is the number of chipped mugs in the sample.

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

   • 20 trials, chipped or not, independent (large delivery), p = 0.07 each time
   • X ~ B(20, 0.07)

2. (b) Find the probability that exactly 2 mugs in the sample are chipped.

   • From the calculator: binomial pdf with n = 20, p = 0.07, x = 2
   • P(X = 2) = 0.252

3. (c) Find the probability that at least 3 mugs in the sample are chipped.

   • P(X ≥ 3) = 1 − P(X ≤ 2)
   • From the calculator: binomial cdf with n = 20, p = 0.07, lower = 0, upper = 2
   • P(X ≤ 2) = 0.8390
   • P(X ≥ 3) = 1 − 0.8390 = 0.161

4. (d) Find the probability that between 1 and 4 mugs in the sample are chipped, inclusive.

   • From the calculator: binomial cdf with n = 20, p = 0.07, lower = 1, upper = 4
   • P(1 ≤ X ≤ 4) = 0.755

A bag holds 7 red sweets and 5 yellow sweets. Four sweets are taken out at random, one after another, without being put back. X is the number of red sweets taken.

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

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

2. (b) Find the probability that the second sweet is red, given that the first sweet was red. Compare it with the probability that the first sweet is red.

   • First sweet red: 7/12 = 0.583
   • Second red, given the first was red: 6/11 = 0.545
   • The probabilities differ, so the binomial conditions fail

Each visitor to a website independently signs up with probability 0.3. 25 visitors arrive. X is the number who sign up.

1. (a) Find the probability that at least 5 but fewer than 10 visitors sign up.

   • P(5 ≤ X < 10) = P(5 ≤ X ≤ 9)
   • From the calculator: binomial cdf with n = 25, p = 0.3, lower = 5, upper = 9
   • P(5 ≤ X ≤ 9) = 0.720

2. (b) Find the probability that at least 10 visitors sign up.

   • P(X ≥ 10) = 1 − P(X ≤ 9)
   • From the calculator: binomial cdf with n = 25, p = 0.3, lower = 0, upper = 9
   • P(X ≤ 9) = 0.8106
   • P(X ≥ 10) = 1 − 0.8106 = 0.189

20% of the trees in a forest have a fungus. The fungus spreads from a tree to its neighbours. 15 trees growing next to each other are checked. X is the number with the fungus. Explain why X may not be binomial.

• Binomial trials must be independent
• The fungus spreads between neighbours, so one tree being infected raises the chance for the next
• The trials are not independent, so X may not be binomial