Question

A bag contains 20 mangoes and 30 oranges out of which 50% of both mangoes and oranges are sour. two fruits are taken out of the bag at random. what is the probability that either both are oranges or both are sour?

Answer 1

Given,

Total no of mangoes = 20

Total no of oranges = 30

Out of total, 50% of each fruit is sour.

2 fruits are picked at random.

Let us assume:

P(A) = probability of getting oranges

P(B) = probability of getting sour fruit

P(A∩B) = probability of getting sour oranges

P(A) = 〖30〗_(C_2 )/〖50〗_(C_2 ) 

 = ((30 ×29 ×28!)/(2 ×28!))/((50 ×49 ×48!)/(2 ×48!))

 =  (30 ×29)/(50 ×49) = 870/(50 ×49)

P(B) =  〖25〗_(C_2 )/〖50〗_(C_2 ) 

 = ((25 ×24 ×23!)/(2 ×23!))/((50 ×49 ×48!)/(2 ×48!))  

 = (25 ×24)/(50 ×49)

 = 600/(50 ×49)

P(A∩B) =  〖15〗_(C_2 )/〖50〗_(C_2 ) 

= ((15 ×14 ×13!)/(2 ×13!))/((50 ×49 ×48!)/(2 ×48!))

= 210/(50 ×49)

Prababilty of getting either oranges or either both are sour

= P(A) + P(B)- P(A∩B) (removing sour oranges probability as it has been added twice)

=  870/(50 ×49)+ 600/(50 ×49) - 210/(50 ×49)

= (870+600-210)/(50 ×49) = 1260/2450 

= 126/245