Theory of Expectation :: Problems on Tossing Coins : Probability Distribution

Problem Back to Problems Page
 
Suppose in a game of coin tossing a person, say X will get Rs. 5 if the head turns up and will lose Rs. 4 if tail turns up. If n (number of trails) in which an unbiased coin is tossed are considered, what is the value of his expectation?

Net Answers :
[Expectation: 0.5n ]

Solution  
 

"x" indicate Mr. X's gain

[Since you are required to find the expected gain of X, the variable would represent Mr. X's gain.]

In a single tossing of the coin, the amount won by Mr. X may be

  • + Rs. 5 (winning Rs. 5) if a head turns up
  • − Rs. 4 (losing Rs. 4) if a tail turns up

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

Total no. of possible choices = 2
Probabilty that the coin would show up

  • A Head ⇒ P(Head) =
    1
    2

  • A Tail ⇒ P(TAIL) =
    1
    2

    Therefore, probability that the amount won by Mr. X would be

  • Rs. 5 ⇒ P(X = 5) =
    1
    2
  • − Rs. 4 ⇒ P(X = − 4) =
    1
    2

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

    Calculations for Mean and Standard Deviations

    x P (X = x) px
    [x × P (X = x)]
    x2 px2
    [x2 × P (X = x)]
    − 4
    1
    2
    − 4
    2
    16
    16
    2
    + 5
    1
    2
    5
    2
    25
    25
    2
    Total 1
    1
    2
    41
    2
    = 0.5 = 20.5

    Expected winnings of Mr. X


    ⇒ Expectation of "x"
    ⇒ E (x) = Σ px
    = 0.5
    Variance of Mr. X's winnings
    ⇒ var (x) = E (x2) − (E(x))2
    ⇒ var (x) = Σ px2 − (Σ px)2
    = 20.5 − (0.5)2
    = 20.5 − 0.25
    = 20.25
    Standard Deviation of the number of heads
    ⇒ SD (x) = + Var (x)
    = + 20.25
    = + 4.5

    Let x1, x2, x3, ... represent the amount won by Mr. X on the first throw, second throw, third throw, ....

    Mr. X's expected amount of winning

    • On the First Throw
      ⇒ E (x1) = 0.5
    • On the Second Throw
      ⇒ E (x2) = 0.5
    • On the Third Throw
      ⇒ E (x3) = 0.5
    • ...
    • ...
    • On the nth Throw
      ⇒ E (xn) = 0.5

    Each trial (throwing of the coin) is identical and therefore the expected amount of winning in each trial would be the same

    Therefore, Mr. X's expected winning in "n" trials (throwings) of the coin

    = E (x1) + E (x2) + E (x3) + ... + E (xn)
    = 0.5 + 0.5 + 0.5 + ... n times
    = n × 0.5
    = 0.5 n

    Alternative

    Where all the trials are identical the expected amount of winning in all the trials together is given by

    Number of Trials × Expectation in each trial

    ⇒ Expected amount of winnings in "n" tosses of the coin

    = No. of times the coin is tossed × Expected winnings per toss
    = n × E (x)
    = n × 0.5
    = 0.5n

    Credit : Vijayalakshmi Desu

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