A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 19 randomly selected pens yields no more than two defective pens. (a) Find the probability that this shipment is accepted if 5% of the total shipment is defective. (Use 3 decimal places.)

Given Information:  

Probability of shipment accepted = p = 5%

Probability of shipment not accepted = q = 95%

Total number of pens = n = 19

Required Information:  

Probability of shipment being accepted with no more than 2 defective pens = P( x ≤ 2) = ?  

P( x ≤ 2) = 0.933

Step-by-step explanation:

The given problem can be solved using Bernoulli distribution  which is given by

P(n, x) = nCx pˣqⁿ⁻ˣ  

The probability of no more than 2 defective pens means

P( x ≤ 2) = Probability of 0 defective pen + Probability of 1 defective pen + Probability of 2 defective pens

P( x ≤ 2) = P(0) + P(1) + P(2)

For P(0) we have p = 0.05, q = 0.95, n = 19 and x = 0

P(0) = 19C0(0.05)⁰(0.95)¹⁹

P(0) = (1)(1)(0.377)

P(0) = 0.377

For P(1) we have p = 0.05, q = 0.95, n = 19 and x = 1

P(1) = 19C1(0.05)¹(0.95)¹⁸

P(1) = (19)(0.05)(0.397)

P(1) = 0.377

For P(2) we have p = 0.05, q = 0.95, n = 19 and x = 2

P(2) = 19C2(0.05)²(0.95)¹⁷

P(2) = (171)(0.0025)(0.418)

P(2) = 0.179

Therefore, the required probability is

P( x ≤ 2) = P(0) + P(1) + P(2)

P( x ≤ 2) = 0.377 + 0.377 + 0.179

P( x ≤ 2) = 0.933

P( x ≤ 2) = 93.3%

Therefore, the probability that this shipment is accepted with no more than 2 defective pens is 0.933.

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

practice. practice. practice.