(I) A ton of 20 bulbs contain 4 blemished ones. One bulb is drawn indiscriminately from the part. What is the likelihood that this bulb is damaged?(ii) Suppose the bulb attracted (I) isn’t damaged and isn’t supplanted. Presently one bulb is drawn indiscriminately from the rest. What is the likelihood that this bulb isn’t deficient?
(I) A ton of 20 bulbs contain 4 blemished ones. One bulb is drawn indiscriminately from the part. What is the likelihood that this bulb is damaged?(ii) Suppose the bulb attracted (I) isn’t damaged and isn’t supplanted. Presently one bulb is drawn indiscriminately from the rest. What is the likelihood that this bulb isn’t deficient?

Solution:

(I) Number of damaged bulbs = 4

The absolute number of bulbs = 20

P(E) = (Number of great results/Total number of results)

∴ Probability of getting a damaged bulb = P (deficient bulb) = 4/20 = ⅕ = 0.2

(ii) Since 1 non-damaged bulb is drawn, then, at that point the absolute quantities of bulbs left are 19

Along these lines, the absolute number of occasions (or results) = 19

Number of non-damaged bulbs = 19-4 = 15

Along these lines, the likelihood that the bulb isn’t deficient = 15/19 = 0.789