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.
Yes as long as the length remained the same each time
Just know that you are smart and that you can do this
The first one is b the second one is b too
5x+130=8x+151
subtract 5x from each side
5x-5x+130=8x-5x+151
130 = 3x+151
subtract 151 from each side
130-151 =3x
-21 = 3x
divide by 3
-21/3 = 3x/3
-7 =x
