Question

In a lottery game a player wins $1,000,000 with probability 0.0000005, wins $200,000 with probability0.000002, and wins $30,000 with probability 0.00001. Suppose the player pays $2 to play the game, what is the player’s expected amount of winnings per game? (Include the minus sign if the answer is negative.)

Answer 1

Answer:

$X_1 = 1000000 , P(X_1)= 0.0000005$

$X_2 = 200000 , P(X_2)= 0.000002$

$X_3 = 30000 , P(X_3)= 0.00001$

$X_4 = -2, P(X_4) = 1-0.0000005-0.000002-0.00001= 0.9999875$

And then replacing we have this for the expected value:

$E(X) = 1000000*0.0000005 +200000*0.000002+30000*0.00001 2*0.9999875$

And after do the math we got:

$E(X) = -0.79998$

And then the expected value for this game is -0.79998 a negative result means NOT a win.

Step-by-step explanation:

For this case we can define the random variable X as the amount that we can win. We can find the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And we have the following probabilities and possible values given:

$X_1 = 1000000 , P(X_1)= 0.0000005$

$X_2 = 200000 , P(X_2)= 0.000002$

$X_3 = 30000 , P(X_3)= 0.00001$

$X_4 = -2, P(X_4) = 1-0.0000005-0.000002-0.00001= 0.9999875$

And then replacing we have this for the expected value:

$E(X) = 1000000*0.0000005 +200000*0.000002+30000*0.00001 2*0.9999875$

And after do the math we got:

$E(X) = -0.79998$

And then the expected value for this game is -0.79998 a negative result means NOT a win.