Question

The probability of getting an item
defective is 0.005. Find the probability
that exactly 3 items in a sample of
200 are defective.
1
(Given e
= 0.3679)
0.5131
0.7131
0.06131
0.4131​

Answer 1

Given : The probability of getting an item defective is 0.005.

To Find  : the probability  that exactly 3 items in a sample of 200 are defective.

Solution:

Using Normal Distribution :

The probability of getting an item defective is 0.005

p = 0.005

1 - p = 1 - 0.005 = 0.995

n = 200

x= 3

P(x) = ⁿCₓpˣ(1-p)ⁿ⁻ˣ

=> P(x) = ²⁰⁰C₃(0.005)³(0.995)¹⁹⁷

=> P(x) =  0.06116

the probability  that exactly 3 items in a sample of 200 are defective. is 0.06116

or using position distribution

np = 200 * 0.005 = 1  = Mean

λ = 1

P(x)  = λˣ  e^(-λ) / x!

P(3) = 1³ e⁻¹/3!  

= e⁻¹/6

= 0.3679/6

= 0.061316

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Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

The probability of getting an item defective is 0.005. Find the probability that exactly 3 items in a sample of 200 are defective.

( Given ${e}^{ - 1}$ = 0.3679 )

• 0.5131

• 0.7131

• 0.06131

• 0.4131

CONCEPT TO BE IMPLEMENTED

POISSON DISTRIBUTION :

X is a poisson random variable with parameter μ then

$\displaystyle \sf{P(X = r) = {e}^{ - \mu}. \frac{ { \mu}^{r} }{r \: !} }$

Where μ > 0 and r = 1 , 2 , 3 , ....

EVALUATION

Here it is given that the probability of getting an item defective is 0.005

Thus p = 0.005

We have to find the probability that exactly 3 items in a sample of 200 are defective

Thus n = Sample size = 200

Since sample size is large

So application of Poisson distribution is recommended

Mean = μ = np = 200 × 0.005 = 1

Also r = 3

Hence the required probability

= P( X = 3 )

$\displaystyle \sf{ = {e}^{ - 1}. \frac{ {1}^{3} }{3 \: !} }$

$\displaystyle \sf{ = \frac{ {e}^{ - 1} }{6} }$

$\displaystyle \sf{ = \frac{ 0.3679 }{6} }$

= 0.06131

FINAL ANSWER

Hence the correct option is 0.06131

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