Of the students in a college, it is known that 60% reside in hostel an

1.	Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hosteler?
Answer» Let E₁ : the event that the student is residing in hostel and E₂ : the event that the student is not residing in the hostel. Let E : a student attains A grade, Then, E₁and E₂ are mutually exclusive and exhaustive. Moreover, P(E₁) = 60% = 60/100 = 3/5 and P(E₂) = 40% = 40/100 = 2/5 Then P(E/E₁) = 30% = 30/100 = 3/10 and P(E/E₂) = 20% = 20/100 = 2/10 By using Baye’s theorem, we obtain P(E₁/E) = (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂)) = (3/5 x 3/10)/(3/5 x 3/10 + 2/5 x 2/10) = 9/(9 + 4) = 9/13

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hosteler?

Answer»

Let E₁ : the event that the student is residing in hostel and E₂ : the event that the student is not residing in the hostel.

Let E : a student attains A grade,

Then, E₁and E₂ are mutually exclusive and exhaustive. Moreover,

P(E₁) = 60% = 60/100 = 3/5 and P(E₂) = 40% = 40/100 = 2/5

Then P(E/E₁) = 30% = 30/100 = 3/10 and P(E/E₂) = 20% = 20/100 = 2/10

By using Baye’s theorem, we obtain

P(E₁/E) = (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂))

= (3/5 x 3/10)/(3/5 x 3/10 + 2/5 x 2/10) = 9/(9 + 4) = 9/13