MATH SOLVE

3 months ago

Q:
# A modified roulette wheel contains 44 numbers, of which 21 are red, 21 are black, and 2 are green. When the roulette wheel is spun, the ball is equally likely to land on any of the 44 numbers. For a bet on black, the house pays 1 to 2 odds. What should the odds actually be to make the bet fair? (Hint: To make the bet fair, the odds paid by the house should be the odds against the ball landing on black.)

Accepted Solution

A:

Answer:To make the bet fair, the odds paid by the house should be 1 to 1.91.Step-by-step explanation:The roulette wheel contains 44 numbers, of which:21 are red.21 are black.2 are green.There are 44 numbers, of which 21 are black, and 44-21 = 23 are not black.To find the odds against the ball landing on black, we have to solve the following direct(cross multiplication) rule of three:1 is to x what 23 is to 44. So:1 - x23 - 44[tex]23x = 44[/tex][tex]23x = 44[/tex][tex]x = \frac{44}{23}[/tex][tex]x = 1.91[/tex]So:To make the bet fair, the odds paid by the house should be 1 to 1.91.