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Question 38 (Case Based Questions) Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage -II contains 6 parrots. One day Arka forgot to lock both cages and two birds flew from Cage-I to Cage-II (simultaneously). Then two birds flew back from cage-II to cage-I(simultaneously). Assume that all the birds have equal chances of flying. Question 38 (i) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I then find the probability that the owl is still in Cage-I. There is only 1 owl, and it’s initially in Cage I. Our goal is to find the probability that after the described movements, the owl is still in Cage I. There are two situations: Situation 1 - The owl stays in Cage I during the first move and remains there. Situation 2 - The owl flies to Cage II in the first move but returns to Cage I in the second move. First, let’s find total number of cases Total cases Here, We choose 2 birds out of 6 birds from Cage 1 Since 2 birds have flown to Cage 2, Cage 2 has 6+2 = 8 birds now We choose 2 birds out of 8 birds from Cage 2 Total cases = 6C2 × 8C2 = 6!/2!4! × 8!/2!6! = (6 × 5)/2 × (8 × 7)/2 = 15 × 28 = 420 Cases involving owl stays in Cage I There are two situations: Situation 1 - The owl stays in Cage I during the first move and remains there. Situation 2 - The owl flies to Cage II in the first move but returns to Cage I in the second move. Situation 1 The owl stays in Cage I during the first move and remains there. Here, Since owl stays in Cage 1, we do not choose it We choose 2 birds out of remaining 5 birds from Cage 1 Since 2 birds have flown to Cage 2, Cage 2 has 6+2 = 8 birds now We choose 2 birds out of 8 birds from Cage 2 Number of ways for Situation 1 = Choosing 2 birds out of remaining 5 birds × Choosing 2 birds out of 8 birds = 5C2 × 8C2 = 5!/2!3! × 8!/2!6! = 10 × 28 = 280 Situation 2 The owl flies to Cage II in the first move but returns to Cage I in the second move. Here, Since owl is flying from Cage 1, we choose 1 more bird We choose 1 birds out of remaining 5 birds from Cage 1 Since 2 birds have flown to Cage 2, Cage 2 has 6+2 = 8 birds now Since owl is flying back We choose 1 bird out of remaining 7 birds from Cage 2 Number of ways for Situation 2 = Choosing 1 birds out of remaining 5 birds × Choosing 1 birds out of 7 birds = 5C1 × 7C1 = 5 × 7 = 35 Required probability Required probability that the owl is still in Cage I = (𝟐𝟖𝟎+𝟑𝟓)/𝟒𝟐𝟎 = 315/420 = 𝟑/𝟒

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.