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  • Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table: One carton was selected at random. What is the probability that it has (i) no defective bulb?

Bulbs are packed in cartons each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:

Number of defective bulbs

0

1

2

3

4

5

6

more than 6

Frequency

400

180

48

41

18

8

3

2

One carton was selected at random. What is the probability that it has

(i) no defective bulb?

(ii) defective bulbs from 2 to 6?

(iii) defective bulbs less than 4?

Answers (2)

\text{Probability is defined as }=\frac{\text {Favourable outcomes}}{\text {Total number of events}}

Here, total events = total cartons =700

(i) no defective bulb

Favourable outcomes =400

\text{p (cartoon has no defective bulb) =}\frac{400}{700}=\frac{4}{7}

(ii) defective bulbs from 2 to 6 = 2 or 3 or 4 or 5 or 6 defective bulbs  

Favourable outcomes = 48 + 41 + 18 + 8 + 3 = 118

\text{p(defective bulb from 2 to 6)}=\frac{118}{700}=\frac{59}{350}

(iii) defective bulbs less than 4 = defective bulbs equal to 0 or 1 or 2 or 3

Favourable outcomes = 400 + 180 + 48 + 41 = 669

\text{p(defective bulbs less than 4)}=\frac{669}{700}

 

Posted by

infoexpert23

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Probability is defined as =\frac{\text {Favourable outcomes}}{\text {Total number of events}}

Here, total events = total cartons = 700

(i) no defective bulb

Favourable outcomes = 400

p (cartoon has no defective bulb) =\frac{400}{700}=\frac{4}{7}

(ii) defective bulbs from 2 to 6 = 2 or 3 or 4 or 5 or 6 defective bulbs  

Favourable outcomes = 48 + 41 + 18 + 8 + 3 = 118

p(defective bulb from 2 to 6) =\frac{118}{700}=\frac{59}{350}

(iii) defective bulbs less than 4 = defective bulbs equal to 0 or 1 or 2 or 3

Favourable outcomes = 400 + 180 + 48 + 41 = 669

p(defective bulbs less than 4)=\frac{669}{700}

 

Posted by

infoexpert23

View full answer

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