Binomial probabilities with the calculator Problem set B One question at a time

\(X \sim B(22, 0.35)\). Find

1. (a) \(P(X = 9)\)

   • From the calculator: binomial pdf with n = 22, p = 0.35, x = 9
   • P(X = 9) = 0.145

2. (b) \(P(X \ge 11)\)

   • P(X ≥ 11) = 1 − P(X ≤ 10)
   • From the calculator: binomial cdf with n = 22, p = 0.35, lower = 0, upper = 10
   • P(X ≤ 10) = 0.8930
   • P(X ≥ 11) = 1 − 0.8930 = 0.107

3. (c) \(P(X < 6)\)

   • P(X < 6) = P(X ≤ 5)
   • From the calculator: binomial cdf with n = 22, p = 0.35, lower = 0, upper = 5
   • P(X ≤ 5) = 0.163

\(X \sim B(5, 0.6)\). Complete the table of the probability distribution of X. Give each value to 3 significant figures.

• X ~ B(5, 0.6)
• From the calculator: binomial pdf with n = 5, p = 0.6, x left blank gives P(X = 0) to P(X = 5)
• x = 0 to 5: 0.0102, 0.0768, 0.230, 0.346, 0.259, 0.0778
• The probabilities add up to 1, apart from rounding

Each of the 48 trees in an orchard independently bears fruit this year with probability 0.35. X is the number of trees that bear fruit.

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

   • 48 trials, success or not, independent, p = 0.35
   • X ~ B(48, 0.35)

2. (b) Find the mean of X.

   • E(X) = np = 48 × 0.35 = 16.8

A printer jams on 15% of mornings, independently from day to day. Over 30 mornings, X is the number of mornings it jams. Find the probability that the printer jams on

1. (a) exactly 4 mornings

   • X ~ B(30, 0.15)
   • From the calculator: binomial pdf with n = 30, p = 0.15, x = 4
   • P(X = 4) = 0.203

2. (b) at least 7 mornings

   • P(X ≥ 7) = 1 − P(X ≤ 6)
   • From the calculator: binomial cdf with n = 30, p = 0.15, lower = 0, upper = 6
   • P(X ≤ 6) = 0.8474
   • P(X ≥ 7) = 1 − 0.8474 = 0.153

3. (c) fewer than 3 mornings

   • P(X < 3) = P(X ≤ 2)
   • From the calculator: binomial cdf with n = 30, p = 0.15, lower = 0, upper = 2
   • P(X ≤ 2) = 0.151

A student tries a puzzle game 6 times. Each attempt independently succeeds with probability 0.35. X is the number of successful attempts.

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

   • X ~ B(6, 0.35)
   • From the calculator: binomial pdf with n = 6, p = 0.35, x left blank gives P(X = 0) to P(X = 6)
   • x = 0 to 6: 0.0754, 0.244, 0.328, 0.235, 0.0951, 0.0205, 0.00184
   • The probabilities add up to 1, apart from rounding

2. (b) Find the probability that at least 3 attempts succeed.

   • P(X ≥ 3) = 1 − P(X ≤ 2)
   • From the calculator: binomial cdf with n = 6, p = 0.35, lower = 0, upper = 2
   • P(X ≤ 2) = 0.6471
   • P(X ≥ 3) = 1 − 0.6471 = 0.353

An archer hits the target with each arrow with probability 0.65, independently. She shoots 16 arrows. X is the number of hits.

1. (a) Find the probability that the archer hits the target at least 13 times.

   • P(X ≥ 13) = 1 − P(X ≤ 12)
   • From the calculator: binomial cdf with n = 16, p = 0.65, lower = 0, upper = 12
   • P(X ≤ 12) = 0.8661
   • P(X ≥ 13) = 1 − 0.8661 = 0.134

2. (b) Find the probability that the archer hits the target fewer than 8 times.

   • P(X < 8) = P(X ≤ 7)
   • From the calculator: binomial cdf with n = 16, p = 0.65, lower = 0, upper = 7
   • P(X ≤ 7) = 0.0671

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

   • From the calculator: binomial cdf with n = 16, p = 0.65, lower = 0, upper = 6
   • P(X ≤ 6) = 0.0229, which is less than 0.05
   • From the calculator: binomial cdf with n = 16, p = 0.65, lower = 0, upper = 7
   • P(X ≤ 7) = 0.0671, which is not less than 0.05
   • The largest k is 6