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JEE Advanced Mathematics Probability 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Mathematics Question (2021) — Solution

Question

Consider three sets E 1 = 1 , 2 , 3 , F 1 = 1 , 3 , 4 and G 1 = 2 , 3 , 4 , 5 . Two elements are chosen at random, without replacement, from the set E 1 , and let S 1 denote the set of these chosen elements. Let E 2 = E 1 - S 1 and F 2 = F 1 ∪ S 1 . Now two elements are chosen at random, without replacement, from the set F 2 and let S 2 denote the set of these chosen elements. Let G 2 = G 1 ∪ S 2 . Finally, two elements are chosen at random, without replacement, from the set G 2 and let S 3 denote the set of these chosen elements. Let E 3 = E 2 ∪ S 3 . Given that E 1 = E 3 , let  p be the conditional probability of the event S 1 = 1 , 2 . Then the value of p is

Options

  1. A. 1 5
  2. B. 3 5
  3. C. 1 2
  4. D. 2 5

Answer

A. 1 5

Step-by-step solution

We need to find p , the conditional probability of the event S 1 = 1 , 2 , given  E 1 = E 3 . So,  p = P S 1 ∩ E 1 = E 3 P E 1 = E 3 = P B 1 , 2 P ( B ) P ( B ) = P B 1 , 2 + P B 1 , 3 + P B 2 , 3 P B 1 , 2 →  If  1 , 2  chosen in the beginning  P B 1 , 3 → If 1 , 3  chosen in the beginning P B 2 , 3 → If  2 , 3  chosen in the beginning (i) If 1 , 2  is chosen in the begining, then F 2 = 1 , 2 , 3 , 4 and S 2 must contain 1 and S 3 = 1 , 2 from G 2 = 1 , 2 , 3 , 4 , 5 So,  P B 1 , 2 = 1 3 × 1 × C 1 3 C 2 4 × 1 C 2 5 = 1 3 × 1 2 × 1 10 (ii) If 1 , 3  is chosen in the begining, then  F 2 = 1 , 3 , 4 and S 2 must contain 1  and S 3 = 1 , 3 from G 2 = 1 , 2 , 3 , 4 , 5 So,  P B 1 , 3 = 1 3 × 1 × C 1 2 C 2 3 × 1 C 2 5 = 1 3 × 2 3 × 1 10 (iii) If 2 , 3  is chosen in the begining, then F 2 = 1 , 2 , 3 , 4 , then two cases are possible 1.  S 2 may contain 1 and getting S 3 = ( 2 , 3 ) from G 2 = 1 , 2 , 3 , 4 , 5 ) 2.   S 2  does not contain 1 and getting ( 2 , 3 ) from G 2 = 2 , 3 , 4 , 5 ] So,  P B 2 , 3 = 1 3 × C 2 3 × 1 C 2 4 × 1 C 2 4 + 1 × C 1 3 C 2 4 × 1 C 2 5 = 1 3 × 1 2 × 1 6 + 1 2 × 1 10 Now,  p = P B 1 , 2 P B 1 , 2 + P B 1 , 3 + P B 2 , 3 = 1 3 × 1 2 × 1 10 1 3 × 1 2 × 1 10 + 1 3 × 2 3 × 1 10 + 1 3 × 1 2 × 1 6 + 1 2 × 1 10 = 1 5

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