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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:

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