   1. Chapter 13 Class 12 Probability
2. Serial order wise
3. Ex 13.2

Transcript

Ex 13.2, 3 A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the Probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale. Number of oranges the box contains = 15 Number of good oranges = 12 Number of bad oranges = 3 3 randomly selected oranges are drawn Box is approved for sale if all 3 oranges are good Probability that box is approved for sale = Probability that first oranges is good × Probability that second orange is good, given first is good × Probability that third orange is good, given first two are good Let ‘O’ represent a good orange So, P(OOO) = P(O) × P(O|O) × P(OO|O) Now, P(O) = Probability that first orange is good = 12﷮15﷯ = 𝟒﷮𝟓﷯ P(O|O) = Probability that second orange is good, given first is good Total number of oranges left = 14 Number of good oranges left = 11 ∴ P(O|O) = 𝟏𝟏﷮𝟏𝟒﷯ P(OO|O) = Probability that third orange is good, given first two are good Total number of oranges left = 13 Number of good oranges left = 10 ∴ P(OO|O) = 𝟏𝟎﷮𝟏𝟑﷯ So, P(OOO) = P(O) × P(O|O) × P(OO|O) = 4﷮5﷯ × 11﷮14﷯ × 10﷮13﷯ = 44﷮91﷯ ∴ P(OOO) = 𝟒𝟒﷮𝟗𝟏﷯

Ex 13.2 