Hi I need help resolving this problem 2315 Apples are packag

Hi, I need help resolving this problem 2.3-15

Apples are packaged automatically in 3-poundbags. Suppose that 4% of the time the bag of applesweighs less than 3 pounds. If you select bags randomlyand weigh them in order to discover one underweight bag of apples,find the probability that the number of bags that must be selected is (a) At least 20. (b) At most 20. (c) Exactly 20. Copyright | Pearson | Probability and Statistical Inference | rpazmino@theezweb.com | Printed from www.chegg.com

Solution

Probability that bag of apples weighs less than 3 pounds = p = 4% = 0.04

Let X : Number of bags selected to get 1 underweight bag of apples

X follows geometric distribution with p=0.04

a. The probability that the number of bags that must be selected to get

1 underweight bag is atleast 20 = P[X>=20] = 1 - P[X < 20]

= 1 - P[X<=20] + P[X=20] ... (I)

For geometric distribution P[X=k] is given as follows:

P[X=k] = (1-p)^k-1 p

and Cumulative distribution function is given as follows:

P[X <= k] = 1 - (1-p)^k

So P[X=20] = (1- 0.04)^20-1 * 0.04 = 0.0184

P[X<=20] = 1 - (1-0.04)²⁰ = 0.5580

Substitute values of P[X=20] and P[X<=20] in I so we get,

The probability that the number of bags that must be selected to get

1 underweight bag is atleast 20 = P[X >=20] = 1 - P[X<=20] + P[X=20]

= 1 - 0.5580+0.0184

= 0.4604

b. The probability that the number of bags that must be selected to get

1 underweight bag is atmost 20 = P[X<=20] = 1 - (1-p)^k

= 1 - (1-0.04)²⁰

= 0.5580

c. The probability that the number of bags that must be selected to get

1 underweight bag is exactly 20 = P[X=20]

= (1- 0.04)^20-1 * 0.04

= 0.0184