A sack contains 20 apples and 30 peaches out of which 50% of both apples and peaches are sour
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Given,
Total no of apples= 20
Total no of peaches= 30
Out of total, 50% of each fruit is sour.
2 fruits are picked at random.
Let us assume:
P(A) = probability of getting peaches
P(B) = probability of getting sour fruit
P(A∩B) = probability of getting sour apples
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 peaches or
either both are sour
= P(A) + P(B)- P(A∩B)
(removing sour peaches 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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Total no of apples= 20
Total no of peaches= 30
Out of total, 50% of each fruit is sour.
2 fruits are picked at random.
Let us assume:
P(A) = probability of getting peaches
P(B) = probability of getting sour fruit
P(A∩B) = probability of getting sour apples
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 peaches or
either both are sour
= P(A) + P(B)- P(A∩B)
(removing sour peaches 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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