Given:

Number Of Defective Bulbs In A Carton

Frequency

0

400

1

180

2

48

3

41

4

18

5

8

6

3

More Than 6

2

The total number of bulbs (N) = 400 + 180 + 48 + 41 + 18 + 8 + 3 + 2 = 700

(I)

Let E be the event that it has no defective bulbs

Total number of bulbs = 700

The number of favourable outcomes to E = 400

Therefore, P(E) = \( \frac{400}{700} \) \(\Rightarrow \frac{4}{7} \)

(II)

Let F be the event that it has 2 to 6 defective bulbs.

Total number of bulbs = 700

The number of favourable outcomes to F = 48 + 41 + 18 + 8 + 3 = 118

Therefore, P(F) =\( \frac{118}{700} \) \(\Rightarrow \frac{59}{350} \)

(III)

Let G be the event that it has less than four defective bulbs.

Total number of bulbs = 700

The number of favourable outcomes to G = 400 + 180 + 41 = 669

Therefore, P(G) =\( \frac{669}{700} \)

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