Binomial probabilities with the calculator Problem set A Worked solutions

\(X \sim B(18, 0.45)\). Find

1. (a) \(P(X = 8)\)

   • From the calculator: binomial pdf with n = 18, p = 0.45, x = 8
   • P(X = 8) = 0.186

2. (b) \(P(X \ge 10)\)

   • P(X ≥ 10) = 1 − P(X ≤ 9)
   • From the calculator: binomial cdf with n = 18, p = 0.45, lower = 0, upper = 9
   • P(X ≤ 9) = 0.7473
   • P(X ≥ 10) = 1 − 0.7473 = 0.253

3. (c) \(P(X < 7)\)

   • P(X < 7) = P(X ≤ 6)
   • From the calculator: binomial cdf with n = 18, p = 0.45, lower = 0, upper = 6
   • P(X ≤ 6) = 0.226

\(X \sim B(5, 0.25)\). Complete the table of the probability distribution of X. Give each value to 3 significant figures.

• X ~ B(5, 0.25)
• From the calculator: binomial pdf with n = 5, p = 0.25, x left blank gives P(X = 0) to P(X = 5)
• x = 0 to 5: 0.237, 0.396, 0.264, 0.0879, 0.0146, 0.000977
• The probabilities add up to 1, apart from rounding

Each of the 36 plants in a greenhouse independently flowers in May with probability 0.25. X is the number of plants that flower in May.

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

   • 36 trials, success or not, independent, p = 0.25
   • X ~ B(36, 0.25)

2. (b) Find the mean of X.

   • E(X) = np = 36 × 0.25 = 9

A school lift is out of order on 8% of days, independently from day to day. Over a term of 40 school days, X is the number of days the lift is out of order. Find the probability that the lift is out of order on

1. (a) exactly 3 days

   • X ~ B(40, 0.08)
   • From the calculator: binomial pdf with n = 40, p = 0.08, x = 3
   • P(X = 3) = 0.231

2. (b) at least 5 days

   • P(X ≥ 5) = 1 − P(X ≤ 4)
   • From the calculator: binomial cdf with n = 40, p = 0.08, lower = 0, upper = 4
   • P(X ≤ 4) = 0.7868
   • P(X ≥ 5) = 1 − 0.7868 = 0.213

3. (c) fewer than 2 days

   • P(X < 2) = P(X ≤ 1)
   • From the calculator: binomial cdf with n = 40, p = 0.08, lower = 0, upper = 1
   • P(X ≤ 1) = 0.159

A nurse gives 6 patients a medicine. Each patient independently needs a second dose with probability 0.25. X is the number of patients who need a second dose.

1. (a) Complete the table of the probability distribution of X. Give each value to 3 significant figures.

   • X ~ B(6, 0.25)
   • From the calculator: binomial pdf with n = 6, p = 0.25, x left blank gives P(X = 0) to P(X = 6)
   • x = 0 to 6: 0.178, 0.356, 0.297, 0.132, 0.0330, 0.00439, 0.000244
   • The probabilities add up to 1, apart from rounding

2. (b) Find the probability that at least 2 patients need a second dose.

   • P(X ≥ 2) = 1 − P(X ≤ 1)
   • From the calculator: binomial cdf with n = 6, p = 0.25, lower = 0, upper = 1
   • P(X ≤ 1) = 0.5339
   • P(X ≥ 2) = 1 − 0.5339 = 0.466

A basketball player makes each free throw with probability 0.72, independently. She takes 15 free throws. X is the number she makes.

1. (a) Find the probability that the player makes at least 12 free throws.

   • P(X ≥ 12) = 1 − P(X ≤ 11)
   • From the calculator: binomial cdf with n = 15, p = 0.72, lower = 0, upper = 11
   • P(X ≤ 11) = 0.6416
   • P(X ≥ 12) = 1 − 0.6416 = 0.358

2. (b) Find the probability that the player makes fewer than 9 free throws.

   • P(X < 9) = P(X ≤ 8)
   • From the calculator: binomial cdf with n = 15, p = 0.72, lower = 0, upper = 8
   • P(X ≤ 8) = 0.0965

3. (c) Find the largest whole number k such that \(P(X \le k) < 0.05\).

   • From the calculator: binomial cdf with n = 15, p = 0.72, lower = 0, upper = 7
   • P(X ≤ 7) = 0.0338, which is less than 0.05
   • From the calculator: binomial cdf with n = 15, p = 0.72, lower = 0, upper = 8
   • P(X ≤ 8) = 0.0965, which is not less than 0.05
   • The largest k is 7