Question

Books are packed in piles each containing 20 books..thirty five piles were examined for defective books and the result are given the following table <br /><br /> Number of defective books 0 , 1 , 2 , 3 ,4 ,5 ,6 ,above 6 <br /><br /><br /> Frequency 400 , 180 ,48 ,41 ,18 ,8 , 3 , 2 <br /><br /><br />one pile was selected at random ,what is the probability that it has .<br /> 1. No defective book <br /> 2.More than 0 but less than 4 defective books <br /> 3. More than 4 defective books<br />​

Answer 1

Given number of defective books in the piles stacked, find the given probabilities

Explanation:

1. let the total number of books be denoted by 'T',  these books were packed in $35$ piles of $20$ books in each. Hence, we have,                                             $T=20(35)\ \ \ \ \ \ \ ->T=700$      
2. now its given that one pile is selected at random than we have ,                                $P(for\ one\ pile)=\frac{20}{35}$  
3. no defective books-   $n(0)->400$                                                                                                       more than zero but less than four defective books-   $n(0-4)-> 180+48+41=269$                                                           more than four defective books-  $n(more\ than\ 4)->8+3+2=13$
4. now probability of no defective books in the selected pile is given by,                                                                                [tex]P_1=\frac{n(0) }{T} .P(for\ one\ pile) \\\\ ->P_1=\frac{400}{700}.\frac{4}{7} \\ ->P_1=0.33 (approx.)[/tex]        ----->ANSWER
5. now probability of more than zero and less than four defective books in the selected pile is given by,                                                            $P_2=\frac{n(0-4) }{T} .P(for\ one\ pile) \\\\->P_2=\frac{269}{700}.\frac{4}{7} \\->P_2=0.22 (approx.)$     ------>ANSWER
6. now probability of more than four defective books is given by,                        $P_3=\frac{n(more\ than\ 4) }{T} .P(for\ one\ pile) \\\\->P_3=\frac{13}{700}.\frac{4}{7} \\->P_3=0.01 (approx.)$------->ANSWER

Answer 2

Answer: > P3 = 0.01(approx.)

Step-by-step explanation:

let the total number of books be denoted by 'T', these books were packed in 35 piles of 20 books in each. Hence, we have, T= 20(35) -> T = 700

2. now its given that one pile is selected at random than we have,

P(for one pile)=

3. no defective books- n(0)-> 400

more than zero but less than four defective books

n(0-4)-> 180+48 +41 269

than four defective books- n(more than 4)->8+3+2=13

4. now probability of no defective books in the selected pile is given by,

PP(for one pile)

> P₁ = 700 7 400 4

-> P₁ = 0.33(approx.)

ANSWER

5. now probability of more than zero and less than four defective books in

the selected pile is given by,

P2().P(for one pile)

269 4 -> P2 = 260-4 700

-> P2 = 0.22(approx.)

------>ANSWER

6. now probability of more than four defective books is given by,

P3: n(more than 4) P(for one pile)

13 -> P3 = 30-4

> P3 = 0.01(approx.)

->ANSWER

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