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) = ?
Answer:
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.
