It is known that the probability that an item produced by a certain machine will be defective is 0.05. If the produced item was sent to the market in packets of 20, find the number of packets containing at least exactly and at most 2 defectives items.
Answers
Answer:
GIVEN: probability of an item produced by a certain machine will be defective is 0.05. the produced items are sent to the market in packets of 20,
TO FIND: number of packets containing at least, exactly and at most 2 defective items
p defective = 0.05
q non defective = 1 - 0.05 = 0.95
packets of 20
=> n = 20
Exactly 2 defective
x = 2
P (x) = ⁿCₓpˣqⁿ⁻ˣ
=> p(2) = ²⁰C₂(0.05)²(0.95)¹⁸ = 0.1886768
188.67
at most 2 defective items
= P(0) + P(1) + P(2)
= ²⁰C₀(0.05)⁰(0.95)²⁰ + ²⁰C₁(0.05)¹(0.95)¹⁹ + ²⁰C₂(0.05)²(0.95)¹⁸
= 0.3584859 + 0.3773536 + 0.1886768
= 0.9245163
= 924.5
ontaining at least 2 defective items
= 1 - P(0) - P(1)
= 1 - 0.3584859 - 0.3773536
= 0.2641605
264 defectives
Poisson approximation to binomial distribution
P(x) = e^{-\lambda} \dfrac{{\lambda}^x}{x!}
Here λ = 20(0.05) = 1
Exactly 2 defective
p(2) = e⁻¹(1)²/2! = 0.1839397
184 defectives
at most 2 defective items
= P(0) + P(1) + P(2)
= e⁻¹(1)⁰/0! + e⁻¹(1)¹/1! + e⁻¹(1)²/2!
= 0.36787944 + 0.36787944 + 0.1839397
= 0.91969858
= 920 defectives
containing at least 2 defective items
= 1 - P(0) - P(1)
= 1 - 0.36787944 - 0.36787944
= 0.26424112
264
Step-by-step explanation: