Question

On an average 5 % items supplied by manufacturer X. are defectives. If a batch of 10 items is inspected: what is the probability that 2 items are
defective​

Answer 1

by using Poisson ratio

y is lamda

Step-by-step explanation:

p(x)=[ (e^-y) * (y^x) ] ÷[x!]

y=[5/100]*10

y=0.5

p(x)=[(e^-0.5) * (0.5^2) ] ÷[2!]

0.075816

Answer 2

Probability that 2 items  of the inspected items are defective is              $(\frac{9}{80} )$ $(\frac{19}{20} )^{8}$ .

Step-by-step explanation:

Given: On an average 5 % items supplied by manufacturer X are defectives.

To Find: Probability that 2 items of the inspected items are defective.

Formula Used:

P(X=k)= nCk  $p^{k}$ $q^{n-k}$        -------------  formula no.01

Here, P(X=k) = The probability of  getting k items  defective  out of  number of n items inspected

            n = number of items inspected

            X= A  random variable which takes value of k

            k = number  of  getting  defective items

            p = probability of getting  one defective  item.

            q = probability of not getting defective item

Solution:

As given on an average 5 % items supplied by manufacturer X. are defectives.

 p = probability of getting  one defective  item out of 100

= 5/100 = 1/20

q = probability of not getting defective item

q = 1-p

   =1-1/20 =19/ 20

n  = 10

k= 2

Substituting the value of n,k,p and q in the formula no.01:

P(X=k)=nCk $p^{k}$ $q^{n-k}$  

=  P(X=2) =  10C2  $(\frac{1}{20} )^{2}$  $(\frac{19}{20} )^{10-2}$

                =  $\frac{10!}{(10-2)! .2!}$  $(\frac{1}{20} )^{2}$ $(\frac{19}{20} )^{8}$  

                =  $\frac{10.9.8!}{8! .2!}$ $(\frac{1}{20} )^{2}$  $(\frac{19}{20} )^{8}$

                = $\frac{45}{1}$ $(\frac{1}{400} )$  $(\frac{19}{20} )^{8}$

                = $(\frac{45}{400} )$  $(\frac{19}{20} )^{8}$

                = $(\frac{9}{80} )$ $(\frac{19}{20} )^{8}$

Thus, Probability that 2 items  of the inspected items are defective is   $(\frac{9}{80} )$ $(\frac{19}{20} )^{8}$ .