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cestrela7 [59]
3 years ago
10

A production process produces 2% defective parts. a sample of 5 parts from the production is selected. what is the probability t

hat the sample contains exactly two defective parts? use the binomial probability function and show your computations to answer this question.
Mathematics
2 answers:
quester [9]3 years ago
7 0
Possible outcomes: defective/non defective

Applying Binomial distribution equation
p(x) = [N!/x!(N-x)!]*(p^x)(q)^N-x

Where,
p = 2% = 0.02
N = 5
And then,
q = 1-p = 1-0.02 = 0.98
x = 2
N-x = 5-2 = 3

Substituting;
p(2) = [5!/2!(5-2)!]*0.02^2*0.98^3 = 0.00376 = 0.376%
Lana71 [14]3 years ago
4 0

Answer:

Probability of a sample that contains exactly two defective parts is .0037 or .37%

Step-by-step explanation:

As we know if P is the probability of achieving k results in n trials then probability formula is P = \binom{n}{k}p^{K}q^{n-k}

In this formula n = number of trials

                        k = number of success

                        (n-k) = number of failures

                         p = probability of success in one trial

                         q = (1-p) = probability of failure in one trial

In this sum n = 5

                  k = 2

number failures (n-k) = (5-2) = 3

                            p = 2% which can be written as .02

                            q = 98% Which can be written as .98

Now putting these values in the formula

                        P = \binom{5}{2}(.02)^{2}(.98)^{5-2}

                        P = \binom{5}{2}(.02)^{2}(.98)^{3}

                   \binom{5}{2}= 5!/3!2!    

                                              = 5×4×3×2×1/3×2×1×2×1

                                              = 5×2 =10

                                       P = 10×(.02)²×(.98)³

                                          = .0037 or .37%

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