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

Solution

"x" indicate the value of the prize won [Since we are required to find the expected prize, the variable would represent the amount of the prize won] The amount of the prize won would be Rs. 2 if a ticket with Rs. 2 prize is drawn Rs. 5 if a ticket with Rs. 5 prize is drawn ⇒ The values carried by the variable ("x") would be either 2, or ⇒ "X" is a discrete random variable with range = {2, 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" Total number of tickets = 5 with a prize of Rs. 2 each + 3 with a prize of Rs. 5 each = 8 Total Number of possible choices (in drawing a ticket) = Number of ways in which a ticket can be drawn from the total 8 = 8C1 ⇒ n = 8 Probabilty that the ticket drawn would be a ticket with a prize of Rs. 2 ⇒ P(Rs. 2) = 3C1 × 5C0 8C1 Tickets with Rs. 2 Prize Tickets with Rs. 5 Prize Total Available 3 5 8 To Choose 1 0 1 Choices 3C1 5C0 8C1 = 3 × 1 8 = 3 8 with a prize of Rs. 5 ⇒ P(Rs. 5) = 3C0 × 5C1 8C1 Tickets with Rs. 2 Prize Tickets with Rs. 5 Prize Total Available 3 5 8 To Choose 0 1 1 Choices 3C0 5C1 8C1 = 1 × 5 8 = 5 8 Probability for the prize won to be Rs. 2 ⇒ P(X = 2) = P(Rs. 2) = 3 8 Rs. 5 ⇒ P(X = 5) = P(Rs. 5) = 5 8 The probabilty distribution of "x" would be x 2 5 P(X = x) 3 8 5 8 Calculations for Mean and Standard Deviations x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 2 3 8 6 8 4 12 8 5 5 8 25 8 25 125 8 Total 1 31 8 137 8 = 3.875 = 17.125 Expected value of the prize ⇒ Expectation of "x" ⇒ E (x) = Σ px = 3.875 Variance of the prize ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 17.125 − (3.875)2 = 17.125 − 15.02 = 2.105 Standard Deviation of the prize ⇒ SD (x) = + √ Var (x) = + √ 2.105 = + 1.451 Where Two Tickets are Drawn Drawing one ticket and drawing two tickets form two different experiments. "y" indicate the amount of the prize won [Since we are required to find the expected value of the game, the variable would represent the amount of prize won by the man.] The amount of prize won by the man would be Rs. 4 if both the tickets drawn have a prize of Rs. 2 Rs. 7 if one ticket drawn has a prize of Rs. 2 and the other has a prize of Rs. 5 Rs. 10 if both the tickets drawn have a prize of Rs. 5 ⇒ The values carried by the variable ("y") would be either 4, 7 or 10 ⇒ "Y" is a discrete random variable with range = {4, 7, 10} "Y" represents the random variable and P(Y = y) represents the probability that the value within the range of the random variable is a specified value of "y" Total number of Tickets = 3 with a prize of Rs. 2 each + 5 with a prize of Rs. 5 each = 8 Total Number of possible choices (in drawing two tickets) = Number of ways in which the two tickets can be drawn from the total 8 = 8C2 ⇒ n = 8 × 7 2 × 1 = 28 Probabilty that the two tickets drawn would be Both with a prize of Rs. 2 ⇒ P(22) = 3C2 × 5C0 8C2 Tickets with Rs. 2 Prize Tickets with Rs. 5 Prize Total Available 3 5 8 To Choose 2 0 2 Choices 3C2 5C0 8C2 = 3 × 2 2 × 1 × 1 28 = 3 28 A ticket with a prize of Rs. 2 and a ticket with a prize of Rs. 5 ⇒ P(25) = 3C1 × 5C1 8C2 Tickets with Rs. 2 Prize Tickets with Rs. 5 Prize Total Available 3 5 8 To Choose 1 1 2 Choices 3C1 5C1 8C2 = 3 1 × 5 1 28 = 15 28 Both with a prize of Rs. 5 ⇒ P(55) = 3C0 × 5C2 8C2 Tickets with Rs. 2 Prize Tickets with Rs. 5 Prize Total Available 3 5 8 To Choose 0 2 2 Choices 3C0 5C2 8C2 = 1 × 5 × 4 2 × 1 28 = 10 28 Probability for the prize won to be Rs. 4 ⇒ P(Y = 4) = P(22) = 3 28 Rs. 7 ⇒ P(Y = 7) = P(55) = 15 28 Rs. 10 ⇒ P(Y = 10) = P(55) = 10 28 The probabilty distribution of "y" would be x 4 7 10 P(Y = y) 3 28 15 28 10 28 Calculations for Mean and Standard Deviations y P (Y = y) py [x × P (Y = y)] y2 py2 [y2 × P (Y = y)] 4 3 28 12 28 16 48 28 7 15 28 105 28 49 735 28 10 10 28 100 28 100 1000 28 Total 1 217 28 1,783 28 = 7.75 = 63.68 The expected value of the game ⇒ Expectation of "x" ⇒ E (x) = Σ px = Rs. 7.75 Variance of the amount of the prize won ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 63.68 − (7.75)2 = 63.68 − 60.06 = 3.62 Standard Deviation of the amount of the prize won ⇒ SD (x) = + √ Var (x) = + √ 3.62 = + 1.903