Theory of Expectation :: Drawing Balls, Cards, Items, Products

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A bag contains 5 white and 7 black balls. Find the expectation of a man who is allowed to draw two balls from the bag and who is to get one rupee for each black ball and two rupee for each white ball drawn

Net Answers :
[Expectation: 2.83 ; Variance: 0.4617 ; Standard Deviation: +0.679]

Solution  
 

"x" indicate the amount got by the man

[Since we are required to find the mans expectation, the variable would represent the amount got by the man]

The amount got by the man would be

  • Rs. 2 if he draws "0 White and 2 Black" balls
  • Rs. 3 if he draws "1 White and 1 Black" ball
  • Rs. 4 if he draws "2 White and 0 Black" balls

⇒ The values carried by the variable ("x") would be either 2, 3 or 4
⇒ "X" is a discrete random variable with range = {2, 3, 4}

"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"

Total number of balls = 5 White + 7 Black
= 12

Total Number of possible choices (in drawing the two balls)

= Number of ways in which the two balls can be drawn from the total 12
= 12Cs

⇒ n =
12 × 11
2 × 1
= 66

Probabilty that the two balls drawn would be

  • "0 White and 2 Black"

    ⇒ P(0W2B) =
    5C0 × 7C2
    12C2
    White Black Total
    Available 5 7 12
    To Choose 0 2 2
    Choices 5C0 7C2 12C2


    =
    1 ×
    7 × 6
    2 × 1
    66
    =
    21
    66
    =
    7
    22

  • "1 White and 1 Black"

    ⇒ P(1W1B) =
    5C1 × 7C1
    12C2
    White Black Total
    Available 5 7 12
    To Choose 1 1 2
    Choices 5C1 7C1 12C2


    =
    5
    1
    ×
    7
    1
    66
    =
    35
    66

  • "2 White and 0 Black"

    ⇒ P(2W0B) =
    5C2 × 7C0
    12C2
    White Black Total
    Available 5 7 12
    To Choose 2 0 2
    Choices 5C2 7C0 12C2


    =
    5 × 4
    2 × 1
    × 1
    66
    =
    10
    66

    Probability for the amount got by the person to be

  • Rs. 2 ⇒ P(X = 2) = P(0W2B)
    =
    21
    66
  • Rs. 3 ⇒ P(X = 3) = P(1W1B)
    =
    35
    66
  • Rs. 4 ⇒ P(X = 4) = P(2W0B)
    =
    10
    66

    The probabilty distribution of "x" would be
    x 2 3 4
    P(X = x)
    21
    66
    35
    66
    10
    66

    Calculations for Mean and Standard Deviations

    x P (X = x) px
    [x × P (X = x)]
    x2 px2
    [x2 × P (X = x)]
    2
    21
    66
    42
    66
    4
    84
    66
    3
    35
    66
    105
    66
    9
    315
    66
    4
    10
    66
    40
    66
    16
    160
    66
    Total 1
    187
    66
    559
    66
    = 2.83 = 8.47

    The mans expectation


    ⇒ Expectation of "x"
    ⇒ E (x) = Σ px
    = Rs. 2.83
    Variance of the amount got by the man
    ⇒ var (x) = E (x2) − (E(x))2
    ⇒ var (x) = Σ px2 − (Σ px)2
    = 8.47 − (2.83)2
    = 8.47 − 8.0083
    = 0.4617
    Standard Deviation of the amount got by the man
    ⇒ SD (x) = + Var (x)
    = + 0.4617
    = + 0.679

    Credit : Vijayalakshmi Desu

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