Question

a factory produces the same number of good and defective products, 3 are selected randomly and are inspected. what are the odds of obtaining:

1) 2 defective products

2) at most 2 defective products

3) no defective products

4) at least 1 defective product

5) 3 defective products

Answer #1

P(defective product), p = 0.5

P(good product), q = 0.5

Sample size, n = 3

Binomial distribution: P(X) = nCx p^{x} q^{n-x},
where x is the number of defective products

1) P(2 defective products), P(X = 2) = 3C2 x 0.5^{2} x
0.5

= 0.375

Odds of 2 defective products = 0.375 : (1 - 0.375)

= **3 : 5**

2) P(at most 2 defective products) = 1 - P(3 defective)

= 1 - 0.5^{3}

= 0.875

Odds of at most defective products = 0.875 : (1-0.875)

= **7 : 1**

3) P(no defective products) = P(0)

= 0.5^{3}

= 0.125

Odds of no defective products = 0.125 : (1-0.125)

= **1 : 7**

4) P(at least 1 defective product) = 1 - P(no defective product)

= 1 - 0.125

= 0.875

Odds of at least 1 defective product = 0.875 : (1-0.875)

= **7 : 1**

5) P(3 defective products) = 0.5^{3}

= 0.125

Odds of 3 defective products = 0.125 : (1-0.125)

= **1 : 7**

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