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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