37. 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 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 hosteller?

0 2 Views | Posted 4 months ago
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  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    37. Let, E1 :event that the student in hostel

    E2 : event that the student is day scholar

    A : be the event of the chosen student get grade A

    P(E1)=60%=60100=0.6

    P(E2)=40%=40100=0.4

    P(A/E1)=P (student getting an A grade is hostler) =30% =30100=0.3

    P(A/E2)=P ( student getting an A grade is a day scholar) =20%=20100=0.2

    Then, by Baye’s theorem,

    P(E1/A)=P(E1).P(A/E1)P(E1).P(A/E1)+P(E2).P(A/E2)

    =0.6×0.30.6×0.3+0.4×0.2=0.180.26=1826=913

Similar Questions for you

A
alok kumar singh

P (2 obtained on even numbered toss) = k (let)

P (2) = 1 6  

P (  2 ¯ )= 5 6  

k = 5 6 × 1 6 + ( 5 6 ) 3 × 1 6 + ( 5 6 ) 5 × 1 6 + . . .

= 5 6 × 1 6 1 ( 5 6 ) 2

= 5 1 1

A
alok kumar singh

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability  = 3 0 1 1 * 1 1 = 3 0 1 2 1

A
alok kumar singh

P (2W and 2B) = P (2B, 6W) × P (2W and 2B)

+ P (3B, 5W) × P (2W and 2B)

+ P (4B, 4W) × P (2W and 2B)

+ P (5B, 3W) × P (2W and 2B)

+ P (6B, 2W) × P (2W and 2B)

(15 + 30 + 36 + 30 + 15)

           

= 3 6 1 2 6

= 1 8 6 3

= 6 2 1

= 2 7

             

A
alok kumar singh

Let probability of tail is   1 3

Probability of getting head = 2 3  

Probability of getting 2 heads and 1 tail

= ( 2 3 × 2 3 × 1 3 ) × 3

= 4 2 7 × 3

= 4 9                  

                   

                   

V
Vishal Baghel

ax2 + bx + c = 0

D = b2 – 4ac

D = 0

b2 – 4ac = 0

b2 = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

a = 4 c = 1 a = 2 c = 2 a = 1 c = 4 } 3 w a y s

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability = 5 2 1 6   

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