52. Find the probability distribution of:

(i) Number of heads in two tosses of a coin.

(ii) Number of tails in the simultaneous tosses of three coins.

(iii) Number of heads in four tosses of a coin.

0 7 Views | Posted 4 months ago
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  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    (i) When one coin is tossed twice, the sample space is

    {HH, HT, TH, TT}

    Let X represent the number of heads.

    ∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0

    Therefore, X can take the value of 0, 1, or 2.

    It is known that,

    P (HH) = P (HT) = P (TH) = P (TT) = 1/4

    P (X = 0) = P (TT) = 1/4

    P (X = 1) = P (HT) + P (TH) = 14 + 1/4 = 1/2

    P (X = 2) = P (HH) = 1/4

    Thus, the required probability distribution is as follows.

    X

    0

    1

    2

    P (X)

    1/4

    1/2

    1/4

    (ii) When three coins are tossed simultaneously, the sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

    Let X represent the number of tails.

    It can be seen that X can take the value of 0, 1, 2, or 3.

    P (X = 0)

    ...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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