29. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that:

(i) Both balls are red.

(ii) First ball is black and second is red.

(iii) One of them is black and other is red.

0 2 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    29. Total number of balls =18

    Number of red balls =8

    Number of black balls =10

    i. Probability of getting red ball at first draw =818=49

    The red ball is replaced after the first draw

     Probability of getting red ball in second draw =818=49

     Probability of getting both the balls red =49×49=1681

    ii. Probability of getting first black ball =1018=59

    The ball is replaced after first draw

    So, the probability of getting second ball as red =818=49

    Probability of getting first ball black and second ball as red =59×49=2081

    iii. Probability of getting first ball as red =818=49

    The ball is replaced after the first draw

    Probability of getting second ball as black =1018=59

    Therefore, probabili

    ...more

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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