35. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

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    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    35. Number of balls contain in win =5 red

    =5 black

    Let the red ball be drawn in the first attempt

     P (drawing a red ball) =510=12

    By question, if added two red balls to the win, then 7 red balls and 5 black balls contain.

    P (drawing a red ball) =712

    Let a black ball be drawn at first attempt

     P (drawing a black ball) =510=12

    If two black balls are added to the win, then 7 black balls and 5 red balls contain.

    P (drawing a red ball) =512

    Therefore, probability of drawing the second ball as of red colour is

    =(12×712)+(12×512)=12(712+512)=12×1212=12×1=12

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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