Question

two numbers are randomly drawn from first 25 natural numbers what is the probability that their sum is even? Ans and win 100₹ paytm !

Answer 1

Question :-- Two numbers are randomly drawn from first 25 natural numbers ... what is the probability that their sum is even ?

Concept and Formula used :--

→ sum of the two numbers is even. This can happen only for the two cases :----

• If Both the numbers are even .
• If Both the numbers are odd .

[[ if one number is odd and one is even, the sum would not be an even number. ]]

→ To calculate the total outcomes of an event where order of the outcomes does not matter we use combination Formula , that is :---- nCr = n! / r! * (n - r)!

→ Probability of an event = No. of Favourable Outcomes / Total Number of Outcomes.

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Lets First Find Total Number of Outcomes ,

we have to select 2 natural numbers from 25 natural numbers.

So,

→ n = 25, r = 2

→ Total Number of Outcomes = 25! / 2! ( 25-2)!

→ Total Number of Outcomes = 25 * 24 * 23! / 2! * 23!

→ Total Number of Outcomes = 25 * 24 / 2 * 1

→ Total Number of Outcomes = 25 * 12 = 300 ..

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in 25 natural Numbers now, we have ,

➠ Total Even Numbers = 12

➠ Total Odd Numbers = 13 .

As Told above , Either we have to Select Two Even Numbers or we have to Choose to odd numbers to make Sum Even .

So,

Number of favorable choices :---

➲ Number of ways in which the two numbers can be drawn such that both are odd or both are even

➲ Number of ways in which 2 odd numbers can be drawn + Number of ways in which 2 even numbers can be drawn

➲ (Number of ways in which 2 odd numbers can be drawn from the available 13) + (Number of ways in which 2 even numbers can be drawn from the available 12)..

➲ ( n = 13 , r = 2 ) + ( n = 12 , r = 2 )

➲ [ 13! / 2! * (13-2)! ] + [ 12! / 2! * (12-2)! ]

➲ [ 13*12*11! / 2*11! ] + [ 12*11*10! / 2*10! ]

➲ [ 13*12/2 ] + [ 12*11/2 ]

➲ 78 + 66

➲ 144

______________________________

⛬ Required Probability = (Number of favorable choices) /.

(Total Number of Possible Choices )

✰) Required Probability = 144 / 300

✰) Required Probability = ( 12 / 25 )

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❂ Hence, Probability of Selecting two natural Numbers From First 25 natural numbers such that their sum is even number is 12/25 .. ❂