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

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10,000 tickets one sold in a lottery in which there is a first prize of Rs. 5,000 two second prizes of Rs. 1,000 each and ten consolation prizes of Rs. 100 each one ticket cost Rs. 1 find the expected net gain or loss if you buy a ticket.

Net Answers :
[Expectation:− 0.70 ; Variance: 899.91 ; Standard Deviation: +29.999]

Solution  
 

"x" indicate the amount of gain

[Since we are required to find the expected gain, the variable would represent the amount of gain]

Gain = Prize amount − Cost of the ticket

The persons gain would be

  • Rs. 4,999 if his ticket gets a prize of Rs. 5,000
    [Gain = Rs. 5,000 − Rs. 1]
  • Rs. 999 if his ticket gets a prize of Rs. 1,000
    [Gain = Rs. 1,000 − Rs. 1]
  • Rs. 99 if his ticket gets a prize of Rs. 100
    [Gain = Rs. 100 − Rs. 1]
  • − Rs. 1 if his ticket does not get any ticket with a prize
    [Gain = Rs. 0 − Rs. 1]

⇒ The values carried by the variable ("x") would be either − 1, 99, 999 or 4,999
⇒ "X" is a discrete random variable with range = {− 1, 99, 999, 4,999 }

"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 tickets = 1 with a prize of Rs. 5,000 + Two with a prize of Rs. 1,000 + 10 with a prize of Rs. 10 + 9,987 with no prize
= 10,000

Total Number of possible choices (in drawing a ticket)

= Number of ways in which a ticket can be drawn from the total 10,000
= 10,000C1

⇒ n = 10,000

Probabilty that the ticket drawn would be

  • a ticket with a prize of Rs. 5,000

    ⇒ P(Rs. 5,000) =
    1C1 × 2C0 × 10C0 × 9,987C0
    10,000C1
    Tickets with a prize of
    Rs. 5,000 Rs. 1,000 Rs. 100 No Prize Total
    Available 1 2 10 9,987 10,000
    To Choose 1 0 0 0 1
    Choices 1C1 2C0 10C0 9,987C0 10,000C1


    =
    1 × 1 × 1 × 1
    10,000
    =
    1
    10,000

  • a ticket with a prize of Rs. 1,000

    ⇒ P(Rs. 1,000) =
    1C0 × 2C1 × 10C0 × 9,987C0
    10,000C1
    Tickets with a prize of
    Rs. 5,000 Rs. 1,000 Rs. 100 No Prize Total
    Available 1 2 10 9,987 10,000
    To Choose 0 1 0 0 1
    Choices 1C0 2C1 10C0 9,987C0 10,000C1


    =
    1 × 1 × 1 × 1
    10,000
    =
    1
    10,000

  • with No Prize

    ⇒ P(Rs. 0) =
    1C0 × 9,999C1
    10,000C1
    Tickets with
    Rs. 3,000 Prize
    Tickets with
    No Prize
    Total
    Available 1 9,999 10,000
    To Choose 0 1 1
    Choices 1C0 9,999C1 10,000C1


    =
    1 × 9,999
    10,000
    =
    9,999
    10,000

    Probability for the persons income to be

  • Rs. 2,999 ⇒ P(X = 2,999) = P(Rs. 3,000)
    =
    1
    10,000
  • − Rs. 1 ⇒ P(X = − 1) = P(Rs. 0)
    =
    9,999
    10,000

    The probabilty distribution of "x" would be
    x − 1 2,999
    P(X = x)
    9,999
    10,000
    1
    10,000

    Calculations for Mean and Standard Deviations

    x P (X = x) px
    [x × P (X = x)]
    x2 px2
    [x2 × P (X = x)]
    − 1
    9,999
    10,000
    − 9,999
    10,000
    1
    9,999
    10,000
    2,999
    1
    10,000
    2,999
    10,000
    89,94,001
    89,94,001
    10,000
    Total 1
    − 7,000
    10,000
    90,04,000
    10,000
    = − 0.7 = 900.40

    Mean of the persons income
    Expectation of the persons income


    ⇒ Expectation of "x"
    ⇒ E (x) = Σ px
    = − Rs. 0.70
    The man can expect to lose Rs. 0.70

    Variance of the persons income
    ⇒ var (x) = E (x2) − (E(x))2
    ⇒ var (x) = Σ px2 − (Σ px)2
    = 900.40 − (− 0.70)2
    = 900.40 − 0.49
    = 899.91
    Standard Deviation of the persons income
    ⇒ SD (x) = + Var (x)
    = + 899.91
    = + 29.999

    Credit : Vijayalakshmi Desu

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