prob....

1. If 2 no.s x and y are chosen from the set of natural no.s, then probabilty that x^2 +y^2 is divisible by 5 is.....

2. If sum of two positive integers is 2n , then the probabilty that their product is grtr than 3/4th of their maximum product is .....

9 Answers

1
°ღ•๓яυΠ·

1 ,2 ,3 , 4 , 5, 6, 7, 8, 9
1, 4, 9, 16 25,36,49,64,81

so nw u can choose those wer x^2+y^2 wil hav 0 or 5 in d unitz place

(1,2) (1,3)(2,4)(3,9)(4,8).................

they can b like 1 or 2 or say 2 or 1...........

similarly u'll get d asnwer ......

they hav said natural nos wich goes upto infi
bt we can consider a group n then it periodically egts repated so willl b d prob

62
Lokesh Verma ·

1) x odd, y even or y odd, x even

Is the set of natural numbers given? :O

Or is it any natural number?

13
deepanshu001 agarwal ·

x and y are chosen from naturall numbers....

62
Lokesh Verma ·

oh sorry i made a mistake...

I put the numbers in the form 4n+1

:((

you have to take the sets

5n, 5n+1, 5n+2, 5n+3 and 5n+4

squares are

5n, 5n+1, 5n-1, 5n-1 and 5n+1

so now you can do the remaining?

62
Lokesh Verma ·

2. If sum of two positive integers is 2n , then the probabilty that their product is grtr than 3/4th of their maximum product is .....

maximum product is n2

we want the product to be less than 3/4 n2

k x (2n-k) < 3/4 n2

now try

13
deepanshu001 agarwal ·

nao i m gettin 5/9 but sir but wat i did was ...

1,4,9,6,5,0 are possible unit digits for a sq ... so 11/36.... where did i go wrong

13
deepanshu001 agarwal ·

2nd to ho gaya ...... but 1st one mein im even more confused nao.....

62
Lokesh Verma ·

depanshu,

each of those have a different probability of coming..

13
deepanshu001 agarwal ·

ok ....i m gettin it thats bcz each of them hav cum from diff form of the no.s ....

and wat 1 did usmien i counted the same case twice...... 5n

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