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1 year ago

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Jane has a deck of 10 cards numbered 1 through 10. She is playing a game of change This game is this: Jane chooses one card from

the deck at random. She wins an amount of money equal to the value of the card if an odd numbered card is drawn. She loses $4.20 if an even numbered card is drawnA) find expected value of playing the game.B what can Jane expect in the long run, after playing the game many times?(She replaces the card in the deck each time)

1 answer:

nikklg [1K]1 year ago

4 0

we know that

total cards=10

even numbers=2,4,6,8,10 -----> 5 cards

the probability that getting an even number is

P=5/10

The probability that getting an odd number

n 1 -----> P=1/10

n3 ----> P=1/10

n5 ---> P=1/10

n7 ----> p=1/10

n 9 ---> P=1/10

The expected value is equal to

EV=(1)(1/10)+(3)(1/10)+(5)(1/10)+(7)(1/10)+(9)(1/10)-(4.20)(5/10)

EV=(25/10)-(21/10)

EV=4/10

EV=$0.40

<h2>The expected value is $0.40</h2>

Part B

Jane can expect to win money after playing the game many times

the answer is

<h2>Jane can expect to gain money</h2><h2>She can expect to win $0.40 per draw</h2>

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