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BigorU [14]
3 years ago
8

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?
Mathematics
1 answer:
ollegr [7]3 years ago
4 0

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

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Points -2 and -6 on the number line are the two solutions.

Step-by-step explanation:

Use the definition of absolute value as a starting point

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To solve the equation, you need to treat the two cases as above:

|x+4|=x+4=2\,\,\,\mbox{for}\,\,x+4\geq 0\implies x\geq -4\\x+4=2\implies x=-2

The solution x=-2 is consistent with the condition x>=-4, so it is the first and valid solution. Now the second case of the absolute value:

|x+4|=-(x+4)=2\,\,\,\mbox{for}\,\,x+4< 0\implies x

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