Question

Q2. 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​

Answer 1

total no. of mangoes=21+3

                                  =24

a)

No. of defective mangoes=3

  total no. of mangoes=24

P(E)=no. of desired outcomes/no. of total outcomes

     =3/24=1/8=0.125

b)

No. of good mangoes=21

total no. of mangoes=24

P(E)=no. of desired outcomes/no. of total outcomes

      =21/24=7/8=0.875

:)