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Sagot :
Answer:
[tex]E=-\$0.47[/tex]
Step-by-step explanation:
From the question we are told that:
Sample size
Number of Ace [tex]A=4-1=3[/tex]
Number of Face card [tex]F=12[/tex]
Number of other cards [tex]O=36[/tex]
Generally the Probability of drawing an ace is mathematically given by
[tex]P(A)=\frac{A}{n}\\P(A)=\frac{3}{51}\\P(A)=0.06[/tex]
Generally the Probability of drawing an Face card is mathematically given by
[tex]P(F)=\frac{F}{n}\\P(F)=\frac{12}{51}\\P(F)=0.24[/tex]
Generally the Probability of drawing an Face card is mathematically given by
[tex]P(F)=\frac{F}{n}\\P(F)=\frac{36}{51}\\P(F)=0.71[/tex]
Therefore expected value with one ace removed E
[tex]E=0.06*20+0.24*5-0.71*4[/tex]
[tex]E=-\$0.47[/tex]
The expected value of the game with one ace removed is -0.44 and this can be determined by using the concept of probability.
Given :
- Drawing a Cardone ace is removed from the deck.
- For this game, you must draw a card from a standard 52-card deck, with one ace removed.
- If you draw an Ace, you win $20.00. If you draw a face card, you win $5.00.
So from the deck of the 52 cards number of Ace cards are 3, the number of Face cards are 12 and the other cards are 36.
Drawing an Ace card has a probability of:
[tex]\rm P(A) = \dfrac{A}{n}[/tex]
[tex]\rm P(A) = \dfrac{3}{51}[/tex]
P(A) = 0.06
Drawing a Face card has a probability of:
[tex]\rm P(F) = \dfrac{F}{n}[/tex]
[tex]\rm P(F) = \dfrac{12}{51}[/tex]
P(F) = 0.24
Drawing other cards has a probability of:
[tex]\rm P(O) = \dfrac{O}{n}[/tex]
[tex]\rm P(O) = \dfrac{36}{51}[/tex]
P(O) = 0.71
So, the expected value with one Ace removed is:
[tex]\rm P = 0.06\times 20 + 0.24\times 5 - 0.71 \times 4[/tex]
P = 1.2 + 1.2 - 2.84
P = 2.4 - 2.84
P = -0.44
For more information, refer to the link given below:
https://brainly.com/question/23017717
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