Question

A box has 20 pens of which 2 are defective. calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.

Answer 1

= 2/20
= 1/10

= 20-5/2
= 15/2

Answer 2

Answer:

0.99144.

Step-by-step explanation:

Total pens = 20

Defective pens = 2

Probability of getting defective pens = $\frac{2}{20}=\frac{1}{10}$

Now we are supposed to find the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective

So, we will use binomial

$P(X=r) =^nC_r p^r q^{n-r}$

p = Probability of success

q is the probability of failure

in this case success is getting defective pen

So, q =  $\frac{1}{10}$

p = 1-q= $1-\frac{1}{10}=\frac{9}{10}$

we are given that out 5 at most 2 are defective

So, r can be 0,1,2

n = 5

Substitute the value in the formula

$P(X) =^{5}C_{0} (\frac{1}{10}^)0 (\frac{9}{10})^{5-0}+^{5}C_{1} (\frac{1}{10}^)1 (\frac{9}{10})^{5-1}+^{5}C_{2} (\frac{1}{10}^)2 (\frac{9}{10})^{5-2}$

$P(X) =^{5}C_{0}(\frac{9}{10})^{5}+^{5}C_{1} (\frac{1}{10}) (\frac{9}{10})^{4}+^{5}C_{2} (\frac{1}{10})^2 (\frac{9}{10})^{3}$

$P(X) =(\frac{9}{10})^{5}+5(\frac{1}{10}) (\frac{9}{10})^{4}+10(\frac{1}{10})^2 (\frac{9}{10})^{3}$

$P(X) =0.99144$

Thus the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective is 0.99144.