Respuesta :
The probability of selecting no more than 2 defectives in a sample of 10 a) without replacement is 0.36077 and b) with replacement is 0.14093
a)
Here it is given that the objects are selected without replacement one after the other and the box contains only 20 items
Here we can see that the no. of items i.e the population is finite and the items are selected one after another without replacement.
Hence, this is a case of Hyper-Geometric distribution.
For a Hyper-Geometric distribution
P(X = x) = [tex]\frac{^{k}C_x X ^{N-k}C_N-x }{^{N}C_n }[/tex]
where,
N = The Population
n = no. of samples selected
k = no. of success.
Here
k = no. of defectives present
= 20% of 20
= 20/100 X 20
= 4
n = 10
N = 20
x = 2
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 0)
Therefore we get
⁴C₀ X ⁽²⁰ ⁻ ⁴⁾C₍₁₀ ₋ ₀₎ / ²⁰C₁₀ + ⁴C₁ X ⁽²⁰ ⁻ ⁴⁾C₍₁₀ ₋ ₁₎ / ²⁰C₁₀ +
⁴C₂ X ⁽²⁰ ⁻ ⁴⁾C₍₁₀ ₋ ₂₎ / ²⁰C₁₀
= ⁴C₀ X ¹⁶C₁₀ / ²⁰C₁₀ + ⁴C₁ X ¹⁶C₉ / ²⁰C₁₀ + ⁴C₂ X ¹⁶C₈ / ²⁰C₁₀
ᵇCₐ = b!/(b - a)! X a!
Therefore,
⁴C₀ X ¹⁶C₁₀ / ²⁰C₁₀ + ⁴C₁ X ¹⁶C₉ / ²⁰C₁₀ + ⁴C₂ X ¹⁶C₈ / ²⁰C₁₀
= 0.04334 + 0.24768 + 0.06975
= 0.36077
b)
Here the sample is selected with replacement
There are 4 defective and 16 non-defective items
Let A be the event of selecting not more than 2 defectives with replacement
According to the problem, 2 out of 4 defectives are selected while the rest are not defective.
Hence n(A) = n(getting 0 defective) + n(getting 1 defective) + n_getting 2 defectives)
= 16¹⁰ + 4 X 16⁹ + 4² X 16⁸
= 1443109011456
Similarly, the total no. of ways to choose an item is
n = 20¹⁰
Probability of any event
= no.of favorable outcomes/total no. of outcomes
Hence, P(A) = n(A) / n
= 4² X 16⁸ / 20¹⁰
= 0.14093
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