Sarah draws a card from a deck of 52 cards. She receives 40 cents for a heart, 50 cents for an ace, and 90 cents for the ace of hearts. The cost of a draw is 15 cents. How much will Sarah lose if she plays this game?
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ok, first of all, you need to know exactly how many hearts, aces, and aces of hearts there are.
Then, to find the probability of her drawing a particular card is P(particular card)=numer of this type of cards in a deck/total number of cards in a deck
so thats the probability of her winning an amount associated with that card
Now, the question asks how much will she lose. Assuming that she only plays once, the only amount she can lose is 15cents as thats the amount it costs to play the game
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so since the number of winning cards in a deck is lower than the other cards, statistically she WILL lose the game. In the long run, she will not make profit. After one single game, statisticaly she will lose 15 cents
tese are just some ideas I have about this question, I am not good in maths myself so I cant fully answer you question, well, atleast I tried =D
Ok you have to at this like there are 4 different outcomes each with a different payout(gain/loss)
1) you draw the ace of hearts. you profit 75 cents (90 - 15).
the probability of drawing this card is 1/52
2) you draw an ace(not hearts). you profit 35 cents (50 - 15)
the probability is 3/52 since there are 3 possible non-heart aces
3) you draw a heart (excluding the ace of hearts). you profit 25
the probability is 12/52, there are 13 hearts minus the ace
4) you lose so it costs you 15 cents or think of it as you profit -15 cents.
the probability is (52-(1+3+12))/52 = 36/52. so however many remaining cards are left.
Notice how all the numbers in the numerators add up to 52 this way the sum of the probabilities equals 1.
Now you have everything you need to find the expected value of playing this game.
multiply each probability by its corresponding profit and then sum up these values.
E(x) = 75(1/52) + 35(3/52) + 25(12/52) -15(36/52)