Question

2. A box contains 3 defective mangoes and 21 good mangoes. One mango is drawn

from the box at random. Find the probability of getting

a) A defective mango

b) a good mango

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Answer 1

Answer:

a) P(defective) = 3/24 = 1/8

b) P(good) = 21 /24 =7/8

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Answer 2

Total number of mangoes = 21 + 3 = 24 mangoes

We know,

P(E) = (No. of favourable outcomes)/(Total number o outcomes)

a) Defective:-

P(E1) = (No. of favourable mangoes)/(Total number of mangoes)

⇒P(E1) = 3/24 = 1/8

b) Good:-

P(E2) = (No. of favourable mangoes)/(Total number of mangoes)

⇒P(E2) = 21/24 = 7/8