Question

a lot of manufactured items contains 20 percent defectives. A sample of 10 items from this lot is chosen at random. what is the probability that the sample conatins at most 3 defective items ?

Answer 1

Step-by-step explanation:

let S be the Sample space

S=(10)

n(S)=10

there are 20 %of defective items

THEREFORE there are 5 defective items

let A be the event that the items chosen is 3 defective items

n(A)=5

p(A) = 5/10

p(A)= 1/2

Answer 2

The probability that the sample contains at most 3 defective items is $(\frac{524}{125} )(\frac{4}{5} )^{7}$.

Step-by-step explanation:

Given:

A lot of manufactured items contains 20 percent defectives.

A sample of 10 items from this lot is chosen at random.

To Find:

The probability that the sample contains at most 3 defective items.

Solution:

Total no. of items =100

No. of defective items =20

No. of safe items =100−20=80

Let p and q be the probability of defective and safe items respectively.

$p=\frac{20}{100} =\frac{1}{5}$

$q=\frac{80}{100} =\frac{4}{5}$

The nos.of items from this lot is chosen at random  n=10

As we know that

$P(X=r)=n_{C}_{r}(p)^{r} (q)^{n-r}$

Therefore,

$P(x\le3)=P(x=0)+P(x=1)+=P(x=2)+P(x=3)$

$P(x\le3)=10_{C}_{0}(\frac{1}{5} )^{0} (\frac{4}{5} )^{10-0}+10_{C}_{1}(\frac{1}{5} )^{1} (\frac{4}{5} )^{10-1}+10_{C}_{2}(\frac{1}{5} )^{2} (\frac{4}{5} )^{10-2}+10_{C}_{3}(\frac{1}{5} )^{3} (\frac{4}{5} )^{10-3}$

$P(x\le3)=10_{C}_{0}(\frac{1}{5} )^{0} (\frac{4}{5} )^{10}+10_{C}_{1}(\frac{1}{5} )^{1} (\frac{4}{5} )^{9}+10_{C}_{2}(\frac{1}{5} )^{2} (\frac{4}{5} )^{8}+10_{C}_{3}(\frac{1}{5} )^{3} (\frac{4}{5} )^{7}$

$P(x\le3)= (\frac{4}{5} )^{10}+ 2 (\frac{4}{5} )^{9}+(\frac{9}{5} ) (\frac{4}{5} )^{8}+(\frac{24}{25} ) (\frac{4}{5} )^{7}$

               $=(\frac{4}{5}) ^{7}[(\frac{4}{5})^{3} +2(\frac{4}{5} )^{2} +(\frac{9}{5} ) (\frac{4}{5} ) +\frac{24}{25} ]$      

                 $=(\frac{4}{5}) ^{7}[\frac{64}{125} +\frac{32}{25} + \frac{36}{25} +\frac{24}{25} ]$    

                $=(\frac{524}{125} )(\frac{4}{5} )^{7}$

Thus,the probability that the sample contains at most 3 defective items is $(\frac{524}{125} )(\frac{4}{5} )^{7}$.

PROJECT CODE#SPJ2