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

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