IIT JEE , AIEEE Forum - Mathematics , Algebra

krishna.aravapalli

let x be a set containing n elements.Two sub sets A&B of x r choosen at random .the probability that

a)AUB=x is (3/4)^n

b)(A~B) U (B~A)=fi is (1/2)^n

c)Aintersection B=fi is (3/4)^n

coolfishamit

The answer is C.

total possible sets = ( nC1 + nC2 + nC3 +.......nCn ) . ( nC1 + nC2 ......nCn)
                            = 4ⁿ
now,
sets having no element common,

nC1 * (n-1C1 + n-1C2........n-1Cn-1) +
nC2 * (n-2C1 + n-2C2........n-2Cn-2) +
.
.
.
.
nCn * (1)

which is = nC1* 2^n-1 +nC2* 2^n-2 + ............nCn * 1
             = (2+1)ⁿ
             = 3ⁿ
so, required probability is (3/4)ⁿ

Work yourself a little to understand, but if that does not help, do write back.
If understood, plz rate me.

krishna.aravapalli

all the 3 r correct ..........
give me the explanation plzzzzzzzzzz

coolfishamit

Hey, I've told u the way to do one. All the other can be done similarly. Do you need any other assistance in understanding that part. (c)

amar.gupta

good work coolfishamit

just small correction...

total possible sets = ( nC0 +nC1 + nC2 + nC3 +.......nCn ) . ( nC0+nC1 + nC2 ......nCn)
                            = 4ⁿ (then only it will be 4ⁿ you have not taken nC0 option)

similarly:

sets having no element common,

nC0 * (nC0 + nC1+..............nCn)   +                 {you have also missed this opt}
nC1 * (n-1C0 + n-1C1 + n-1C2........n-1Cn-1) +
nC2 * (n-2C0 + n-2C1 + n-2C2........n-2Cn-2) +
.
.
.
.
nCn * (1)

which is =nC0*2ⁿ+nC1* 2^n-1 +nC2* 2^n-2 + ............nCn * 1
             = (2+1)ⁿ
             = 3ⁿ

so, required probability is (3/4)ⁿI think you have got the point.

and dear krishna , Procedure will remain same for all the three, so try to find out the remaining two parts.

Feel free to ask if not able to get the ans.

Ask & Discuss Questions with Community & Experts