A is a set containing 'n' elements A subset P of A is chosen at random The set A is - Maths - Probability

Question

A is a set containing 'n' elements A subset P of A is chosen at
random The set A is reconstructed by replacing the elements of P A
subset Q of A is chosen again at random Find the probability
that
(a)P and Q have same number of elements
(b)The number of elements in P is more than the number of elements
in Q
(c)The number of elements in P is just one more than the number of
elements in Q
(d)Q is a subset of P
(e)P union Q contains exactly r elements(1<=r<=n)

Shivam Mishra · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

aWe know that number of subsets of any set having n elements is 2n
It is convenient to imagine that the choices are made from two distinct n elementsets
There are 2n possible ways to select P and 2n possible ways to select Q
HenceNumber of ways to select subsets P and Q=2n×2n=22n=4nFor favorable case, If P is empty then Q is empty
If P as one element then Q has one element OR If P has two elements, thenQ has 2 elements and so on
Favorable ways=nC0×nC0+nC1×nC1+nC2×nC2+
+nCn×nCnNow1+xn=nC0+nC1x+nC2x2+
+nCnxnx+1n=nC0xn+nC1xn-1+nC2xn-2+
+nCnMultiplying above two equations and equationg coefficients of xn we getCoefficient of xn in 1+xn×x+1n=nC0×nC0+nC1×nC1+nC2×nC2+
+nCn×nCn⇒Coefficient of xn in 1+x2n=nC0×nC0+nC1×nC1+nC2×nC2+
+nCn×nCnAs coefficient of xr in 1+xn is nCr, hence we get:Coefficient of xn in 1+x2n=2nCn⇒nC0×nC0+nC1×nC1+nC2×nC2+
+nCn×nCn=2nCn=Favorable cases⇒Required Probabilty=2nCn4n

Hope this information will clear your doubts about this topic

If you have any doubts just ask here on the ask and answer forum
and our experts will try to help you out as soon as possible
Regards