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melamori03 [73]
3 years ago
7

There are 52 cards in a deck, and 13 of them are hearts. Consider the following two scenarios:

Mathematics
1 answer:
Klio2033 [76]3 years ago
8 0

Answer:

The two scenarios differ.

P(Scenario A) = 0.107

P(Scenario B) = 0.265

Step-by-step explanation:

The two scenarios are different because for the second one you are alredy assuming that the first 3 cards are not hearths, and for that reason, B is more likely to happen that A.

For B to happen, you need to notice that since you remove 3 cards from your deck that are not hearts, then your deck has only 49 cards, and 13 of them are hearts. The probability for a heart to show up is, as a result 13/49 = 0.265 because you have 13 favourable cases from 49 possible.

For A, you need the first card to be anything but a heart. Since 13 cards of the deck are herts, 39 are not, and the probability of that hapening is 39/52 = 3/4. After you remove your first card, the probability of the second one not being heart is 38/51, and the probability for the third one is 37/50 (you are removing one favourable case and one case for the total of cases each time). The probability for the fourth card being a heart assuming that the first three are not was calculated before and it gives us a result of 13/49.

Multiplying everything, we obtain that

P(A) = 3/4*38/51*37/50*13/49 = 0.107.

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