 # NCERT Exemplar Solutions Class 10 Mathematics Solutions for Statistics and Probability - Exercise 13.3 in Chapter 13 - Statistics and Probability

A carton of 24 bulbs contain 6 defective bulbs and one bulb is drawn at random. If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

Total bulbs in carton = 24 →  T(E) = 24

Defective bulbs = 6

Number of favourable outcomes of drawing a bulb which is not defective = 24 – 6 = 18 →  F(E) = 18

Probability that bulb is not defective = P(E) = \frac{\mathrm{F}(\mathrm{E})}{\mathrm{T}(\mathrm{E})}=\frac{18}{24}=\frac{3}{4}

According to the question selected bulb is defective and not replaced.

So, the total remaining bulb = 23 →  T'(E) = 23

Number of favourable outcomes of drawing a second defective bulb i.e., F'(E) = 6 – 1 = 5

P^{\prime}(E)=\frac{F^{\prime}(E)}{T^{\prime}(E)}=\frac{5}{23}

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