IIT JEE , AIEEE Forum - Mathematics , Algebra

ridhima

if 2 natural numbers b/w 1 and 30 are chosen
find the probability that x^2 - y^2 is divisible by 3
just give a hint

ridhima

com on

akhil_o

(x-y)(x+y) must be divisible by 3
so (x-y) or (x+y) or both (x-y)(x+y) are divisible by 3

for x-y divisible by 3
m=3k+1,n=3k+1 or m=3k+2,n=3k+2 or m=3k,n=3k

P(A)=1/3*1/3+1/3*1/3+1/3*1/3=1/3

for x+y divisible by 3
m=3k+1,n=3k+2 or m=3k+2, n=3k+1, m=3k and n=3k
=1/3*1/3+1/3*1/3+1/3*1/3=1/3

for both divisible by 3
1/3*1/3=1/9

P(A)UP(B)=P(A)+P(B)-p(A intersection B)
=1/3+1/3-1/9
=5/9

sweet08

out of 1to 30, 10 no.s r of the form---3m

10 r of ------------3m+1

10 r of ------------3m+2

now, x^2-y^2 to be divisible by 3 then,

x--->3m => y---->3m

x----->3m+1 => y->3m+2

so P=10*10+2(10+10)/30*30==1/3

NOTE:NUMERATOR 2 IS MULTIPLIED BCOZ X---->3M+1

OR X----->3M+2 => Y WILL TAKE THE OTHER VALUE

sugeet

i got 2/3

bcos , no. of outcomes =60,

favourable(x-y)divisible by 3=20

+(x+y)divisible by 3= 20

so 20 + 20 =40/60=2/3

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