Question 38 (i) [Case Based] - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

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