Probability Question 372
Question: Consider a set P containing n elements. A subset A or P is drawn and there after set P is reconstructed. Now one more subset B of P is drawn probability of drawing sets A and B so that $ A\cap B $ has exactly one element is
Options:
A) $ {{(3/4)}^{n}}\cdot n $
B) $ n\cdot {{(3/4)}^{n-1}} $
C) $ (n-1)\cdot {{(3/4)}^{n}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ x_{i} $ be any element of set P, we have following possibilities (i) $ x_{i}\in A,x_{i}\in B; $
(ii) $ x_{i}\in A,x_{i}\notin B; $ (iii)
$ x_{i}\notin A,x_{i}\in B; $
(iv) $ x_{i}\notin A,x_{i}\notin B; $
Clearly, the element $ x_{i}\in A\cap B $ if it belongs to A and B both.
Thus out of these 4 ways only first way is favourable.
Now the element that we want to be in the intersection can be chosen in n different ways, Hence required probability is $ n\cdot {{(3/4)}^{n-1.}} $ .
BETA
