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Aliun [14]
3 years ago
14

A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on four numbers pays 8 to 1 (that is, if you bet $1 and one of t

he four numbers you bet comes up, you get back your $1 plus another $8). How much do you expect to win with a $1 bet on four numbers? HINT [See Example 4.] (Round your answer to the nearest cent.)
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

Answer:

Lose $0.05

Step-by-step explanation:

There are 38 possible spots on the roulette wheel (numbers 1 to 36, 0 and 00).

If the player can choose four numbers on single $1 bet, his chances of winning (W) and losing (L) are as follows:

P(W) = \frac{4}{38} \\P(L) = 1-P(W) = 1-\frac{4}{38} \\P(L) = \frac{34}{38}

The expected value of the bet is given by the probability of winning multiplied by the payout ($8), minus the probability of losing multiplied by the bet cost ($1)

EV=\frac{4}{38}*\$8 -\frac{34}{38}*\$1\\EV= -\$0.05

On each bet, the player is expected to lose 5 cents ($0.05).

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Answer: Non of the option.

Step-by-step explanation:

The given equation is

4x^3 - 6x^2 + 15x + 2 = 0

Let's test 1, if it's one of the root. I.e

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If x = 3

4×27 - 6×9 + 15×3 + 2

108 - 54 + 45 + 2 = 101

Since non of them tend to zero, non of the option is a root of the equation

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