Question

In the casino game of roulette, a gambler can bet on which of 38 numbers the ball will land when the roulette wheel is spun. On a $2 bet, a gambler gains $70 (so a net profit of $68) if he or she chooses the winning number, but loses the $2 otherwise. The probability of choosing the winning number is 1/38 and the probability of not choosing the winning number is 37/38. Let X denote the net gain from a roulette spin. What is the expected winnings for one spin?

Answer 1

Answer:

Expected winnings for one spin = 0.1

Step-by-step explanation:

Expected Value = Sum of [(Probability of X)(Value of X)]

E(X) = Σ [P(X).X]

Two Events : X = +68 , P(X) = 1/38 and X = -2 , P(X) = 37/38

E(X) = [(68) (1/38)] + [(-2)(37/38)]

[(68)(0.03)] + [(-2)(0.97)]

= 2.04 - 1.94

= 0.1

Answer 2

The expected winnings for one spin should be -$0.16.

Given information:

On a $2 bet, a gambler gains $70 (so a net profit of $68) if he or she chooses the winning number, but loses the $2 otherwise. The probability of choosing the winning number is 1/38 and the probability of not choosing the winning number is 37/38.

Calculation of the expected winnings:

$= 68 \times \frac{1}{38} - 2 \times \frac{37}{38} \\\\= \frac{-6}{38}$

= -0.158

= 0.16

Learn more about the number here: https://brainly.com/question/14938106