A set A is containing n elements. A subset P of A is chosen at random. The set is reconstructed by replacing the elements of P. A subset Q of A is again chosen at random. The probability that P and Q have no common elements is :

Engineering

Mathematics

Conditional Probability

Question

${(\frac{5}{6})}^{n}$

${(\frac{3}{4})}^{n}$

${(\frac{3}{5})}^{n}$

${(\frac{2}{3})}^{n}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Set A has n elements. Number of subsets of A = 2ⁿ.

Number of ways of choosing a subset from A = 2ⁿC₁ ways = 2ⁿ ways.

So, number of ways of selecting two subsets from A = 2ⁿ × 2ⁿ = 4ⁿ ways. -----------------(1)

P is a subset of A.

Let set P has r elements.

These r elements are chosen from n elements of set A (r varies from 0 to n).

This can be done in ⁿC_r ways.

Now given that P and Q has no common elements, P∩Q = ∅.

Hence the elements of Q has to be chosen from n – r elements.

This can be done in ^{n – r}C₀ + ^{n – r}C₁ +............+ ^{n – r}C_{n – r} ways = 2^{n – r} ways.

Hence P and Q can be chosen in ⁿC_r × 2^{n – r} ways.

Now r can vary from 0 to n.

Hence total number of ways in which P and Q can be chosen such that they dont have any common elements

= \sum_{r = 0}^{n}^{n} C_{r} \times 2^{n - r}

= \sum_{r = 0}^{n}^{n} C_{r} \times 1^{r} \times 2^{n - r}

= (1 + 2)ⁿ = 3ⁿ -------------------------(2) (Using binomial expansion)

Therefore, probability that P and Q have no common elements

= \frac{3^{n}}{4^{n}} = {(\frac{3}{4})}^{n}

(From (1) and (2))

Please subscribe our Youtube channel to unlock this solution.