A sack contains 40 apples and 60 mangoes out of which 50% of both apples and mangoes are sour. Two fruits are taken out of the sack at random. What is the probability that either both are mangoes or both are sour?
Answers
Answer:
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Step-by-step explanation:
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
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
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