Question

If 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random, 2 bolts will be defective ?

Answer 1

0.1536‬ is probability that out of 4 bolts chosen at random, 2 bolts will be defective

Step-by-step explanation:

20% of the bolts produced by a machine are defective

=> Probability of defective p = 20/100 = 0.2

Probability of non defective = 1 - 0.2 = 0.8

Bolts = 4

P(x) = ⁿCₓpˣ(q)ⁿ⁻ˣ

Defective            Probability

0                          ⁴C₀(0.2)⁰(0.8)⁴  = 0.4096

1                          ⁴C₁(0.2)¹(0.8)³  =   0.4096

2                          ⁴C₂(0.2)²(0.8)²  = 0.1536‬

3                          ⁴C₃(0.2)³(0.8)¹  =  0.0256‬

4                          ⁴C₄(0.2)⁴(0.8)⁰  = 0.0016

probability that out of 4 bolts chosen at random, 2 bolts will be defective

=    ⁴C₂(0.2)²(0.8)²  = 0.1536‬

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Answer 2

Answer:

Given 20% of bolts are defective = $\frac{20}{100}$  = $\frac{1}{5}$

said that less than 2 bolts will be defective so P(X<2)

P(X<2) = P( X=0) + P(X=1)

           = 4C0 ($\frac{4}{5}$)^4 +4C1 ($\frac{4}{5}$)^3

                  =  0.8192

Step-by-step explanation: