From a box containing 10 cards numbered 1 to 10, four cards are drawn together. the probability that their sum is even is
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Solution :-
The sum of four numbers will be even if all the dour numbers are even or odd or two are even and two are odd.
There are 10 cards with numbers from 1 to 10 written on them. So, there are 5 odd numbered cards and 5 even numbered cards.
Required Probability = [(5c₄ + 5c₄) + (5c₂ × 5c₂)]/(10c₄)
⇒ [{(5*4*3*2)/(4*3*2*1)} + {(5*4*3*2)/(4*3*2*1)} + {(5*4)/(2*1)} × {(5*4)/(2*1)}]/{(10*9*8*7)/(4*3*2*1)}
⇒ [(120/24) + (120/24) + {(20/2) × (20/2)}]/(5040/24)
⇒ {(5 + 5) + (10 × 10)}/210
⇒ {10 + 100}/210
⇒ 110/210
= 11/21
Answer.
The sum of four numbers will be even if all the dour numbers are even or odd or two are even and two are odd.
There are 10 cards with numbers from 1 to 10 written on them. So, there are 5 odd numbered cards and 5 even numbered cards.
Required Probability = [(5c₄ + 5c₄) + (5c₂ × 5c₂)]/(10c₄)
⇒ [{(5*4*3*2)/(4*3*2*1)} + {(5*4*3*2)/(4*3*2*1)} + {(5*4)/(2*1)} × {(5*4)/(2*1)}]/{(10*9*8*7)/(4*3*2*1)}
⇒ [(120/24) + (120/24) + {(20/2) × (20/2)}]/(5040/24)
⇒ {(5 + 5) + (10 × 10)}/210
⇒ {10 + 100}/210
⇒ 110/210
= 11/21
Answer.
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