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 the subset of P. A sub-set Q of A is again chosen at random. Find the probability that P and Q have no common elements.
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 the subset of P. A sub-set Q of A is again chosen at random. Find the probability that P and Q have no common elements. Correct Answer (3/4)<sup style=" vertical-align: super; margin: 0px; padding: 0px;font-size:10.5px;">n</sup>
Calculation:
⇒ In set P we can have no element i.e.Φ, 1 element, 2 elements, ...... up to n elements. If we have no element in P, we will leave by all the elements, and a number of set Q formed by those elements will have no common element in common with P. Similarly, if there are r elements in P we are left with the rest of (n - r) element to form Q, satisfying the condition that P and Q should be disjoint.
⇒ Hence, the total number of ways in which P and Q are disjoint can be given by = ∑nr = 0 nCr(2)(n - r) = (3)n
⇒ Suppose you have a 5 element set {1, 2, 3, 4, 5} and you formed a subset of 2 elements, which you can do in 5C2 ways. Let's select 1 and 2, then we are left with 3, 4 and 5.
⇒ Now we can form (2)3 = 8 i.e. {Φ}, {3}, {4}, {5}, {3, 4}, {3, 5}, {4, 5}, {3, 4, 5} subsets which will have no common element with the previously chosen subset.
⇒ So the total number of ways in which you can form two disjoint set is 5C0(2)5 + 5C1(2)4 + 5C2(2)3 + 5C3(2)2 + 5C4(2)1 + 5 C5(2)0 = (3)5
⇒ Total number of ways in which we can form P and Q
⇒ nC0(2)n + nC1(2)n + nC2(2)n + … + nCn(2)n = (4)n
⇒ or simply total number of ways = Total number of ways of forming P × total number of ways of forming Q.= 2n× 2n = (4)n.
⇒ So, required probability = (3/4)n
