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Let a die rolled till 2 is obtained. The probability that 2 obtained on even numbered toss is equal to

Option 1 -

5 1 1

Option 2 -

5 6

Option 3 -

1 1 1

Option 4 -

6 1 1

0 2 Views | Posted 3 weeks ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 weeks ago
    Correct Option - 1


    Detailed Solution:

    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

Similar Questions for you

Given set S = {0, 1, 2, 3, ….., 10}. If a random ordered pair (x, y) of elements of S is chosen, then find probability that |x – y| > 5
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 bag contains 8 balls (black and white). If four balls are chosen without replacement then 2W and 2B are found then the probability that number of white and black balls are same in bag is equal to
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 coin is biased such that head has two chances than tails, what is the probability of getting 2 heads and 1 tail?
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                  

                   

                   

The coefficients a, b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is:
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   

When a missile is fired from a ship, the probability that it is intercepted is 1 3 and the probability that the missile hits the target, given that it is not intercepted, is 3 4 .  If three missiles are fired independently from the ship, then probability that all three hit the target, is:
V
Vishal Baghel

Let

A : Missile hit the target

B : Missile intercepted

P (B) = 1 3 P ( A / B ¯ ) = 3 4

P ( B ¯ ) = 2 3

⇒ P ( B ¯ ∩ A ) = 3 4 × 2 3 = 1 2

Required Probability = 2 3 × 3 4 × 2 3 × 3 4 × 2 3 × 3 4 = 1 8

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