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question 21 (5 points) a dice game involves throwing three dice and bet…

Question

question 21 (5 points)
a dice game involves throwing three dice and betting on one of the six numbers that are on the
dice. the game costs $4 to play, and you win if the number you bet appears on any of the dice. the
distribution for the outcomes of the game (including the profit) is shown below:
number of dice with your number profit probability of observing
0 -$4 125/216
1 $4 75/216
2 $6 15/216
3 $12 1/216
find your expected profit from playing this game.
$4.17
-$0.46
$0.50
$2.36

Explanation:

Step1: Recall the formula for expected value

The formula for the expected value \(E(X)\) of a discrete random variable is \(E(X)=\sum_{i}x_{i}P(x_{i})\), where \(x_{i}\) are the possible values and \(P(x_{i})\) are their corresponding probabilities.

Step2: Calculate each term in the sum

  • For \(x = - 4\) and \(P(x)=\frac{125}{216}\), the term is \((-4)\times\frac{125}{216}=-\frac{500}{216}\)
  • For \(x = 4\) and \(P(x)=\frac{75}{216}\), the term is \(4\times\frac{75}{216}=\frac{300}{216}\)
  • For \(x = 6\) and \(P(x)=\frac{15}{216}\), the term is \(6\times\frac{15}{216}=\frac{90}{216}\)
  • For \(x = 12\) and \(P(x)=\frac{1}{216}\), the term is \(12\times\frac{1}{216}=\frac{12}{216}\)

Step3: Sum up the terms

$$ LATEXBLOCK0 $$

Answer:

\(-\$0.46\)