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

scenario a: four cards are dealt, one at a time, off the top of a well-shuffled deck. what is the probability that a heart turns up on the fourth card, but not before?
scenario b: a deck of cards is shuffled. you have to deal one card at a time until a heart turns up. you have dealt 3 cards, and still have not seen a heart. what is the probability of getting a heart on the 4th card?
calculate the probabilities of the two scenarios and show your work. are the two scenarios different? explain.

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.

you're hilarious but what the hell