Theory of Expectation :: Problems on Throwing/Rolling Dice : Probability Distribution

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A six faced dice is tossed. If a prime number occurs sumeet wins that many number of rupees. But, if a non–prime number occurs he loses that many number of rupees. Determine whether the game is favourable or unfavorable to the player.

Net Answers :
[Expectation:− 0.167 ; Variance: 15.142 ; Standard Deviation: +3.891]

Solution  
 

Let "x" indicate the amount that sumeet wins

[Since you are required to find the expected earning of sumeet, the variable would represent sumeets earnings]

The amount earned by sumeet would be

  • − Rs. 1 if the dice shows up ONE
  • + Rs. 2 if the dice shows up TWO
  • + Rs. 3 if the dice shows up THREE
  • − Rs. 4 if the dice shows up FOUR
  • + Rs. 5 if the dice shows up FIVE
  • − Rs. 6 if the dice shows up SIX
Prime Numbers from 1 to 6 ⇒ {2, 3, 5}
Non- Prime Numbers from 1 to 6 ⇒ {1, 4, 6}

⇒ The values carried by the variable ("x") would be either − 1, + 2, + 3, − 4, + 5 or − 6
⇒ "X" is a discrete random variable with range = {− 6, − 4, − 1, + 2, + 3, + 5}

"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"

In the experiment of tossing a dice, there are six possible elementary events, the events of the die showing up either ONE, TWO, THREE, FOUR, FIVE or SIX all of which are mutually exclusive, equally likely and exhaustive.

Therefore the probability of occurance of each elementary event is 1/6 Probabilty that the dice would show up

  • ONE ⇒ P(ONE) =
    1
    6
  • TWO ⇒ P(TWO) =
    1
    6
  • THREE ⇒ P(THREE) =
    1
    6
  • FOUR ⇒ P(FOUR) =
    1
    6
  • FIVE ⇒ P(FIVE) =
    1
    6
  • SIX ⇒ P(SIX) =
    1
    6

    Probability for the amount earned by sumeet to be

  • − 1 ⇒ P(X = − 1) =
    1
    6
  • + 2 ⇒ P(X = + 2) =
    1
    6
  • + 3 ⇒ P(X = + 3) =
    1
    6
  • − 4 ⇒ P(X = − 4) =
    1
    6
  • + 5 ⇒ P(X = + 5) =
    1
    6
  • − 6 ⇒ P(X = − 6) =
    1
    6

    The probabilty distribution of "x" would be
    x − 1 + 2 + 3 − 4 + 5 − 6
    P(X = x)
    1
    6
    1
    6
    1
    6
    1
    6
    1
    6
    1
    6

    Calculations for Mean and Variance

    x P (X = x) px
    [x × P (X = x)]
    x2 px2
    [x2 × P (X = x)]
    − 1
    1
    6
    − 1
    6
    1
    1
    6
    + 2
    1
    6
    2
    6
    4
    4
    6
    + 3
    1
    6
    + 3
    6
    9
    9
    6
    − 4
    1
    6
    − 4
    6
    16
    16
    6
    + 5
    1
    6
    + 5
    6
    25
    25
    6
    − 6
    1
    6
    − 6
    6
    36
    36
    6
    Total 1
    − 1
    6
    91
    6
    = − 0.167 = + 15.17

    Expectation of the persons gain


    ⇒ Expectation of "x"
    ⇒ E (x) = Σ px
    = − 0.167
    Sumeet can expect to lose Rs. 0.167 Variance of sumeets gain
    ⇒ var (x) = E (x2) − (E(x))2
    ⇒ var (x) = Σ px2 − (Σ px)2
    = 15.17 − (− 0.167)2
    = 15.17 − 0.028
    = 15.142
    Standard Deviation of sumeets gain
    ⇒ SD (x) = + Var (x)
    = + 15.142
    = + 3.891

    Credit : Vijayalakshmi Desu

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