Cards are dealt from a well-shuffled standard deck until th first heart appears. What is the probability...?

Question

A) What is the probability that exactly 5 deals are required?
B) What is the probability that 5 or fewer deals are required?
C) What is the probability that exactly 3 deals were required, given that 5 or fewer were required?
I assumed A) was 13/48 using the conditional probability
P(5th is heart | first 4 are not hearts) = P (5th is & 1,2,3,4 are not) / P(1,2,3,4 are not) and those cancel so it equals P(5th).
I am not sure if I am doing this right, so I need more help.

Answer 1

A) you will get 4 non-hearts and the 5th a heart:
There are 13 hearts and therefore 39 non-hearts.

P(5)=(39/52)* (38/51)* (37/50)* (36/49)* (13/48)

B) P(1) = 13/52 = 1/4
P(2) = (39/52)(13/51)
P(3) = (39/52)(38/51)(13/50)
P(4) = (39/52)(38/51)(37/50)(13/49)
P(5) is above

Your answer is the sum

C) = [P(3 deals AND #deals less than 5)*P(3 deals)]/P(#deals less than 5)

= P(3 deals)/Answer from B

Answer 2

Assuming a standard deck of 52 playing cards, 13 of these are hearts (ace through 10, Jack, Queen, King). So 39 of them are not hearts. The probability of picking a non-heart card first is 39/52. The probability of picking another non-heart card, given that one has already been chosen, is 38/51 (there's one less non-heart card in the deck, which also means one less card in general in the deck). So continuing on like this, the probability of picking 4 non-heart cards in a row is (39/52) * (38/51) * (37/50)*(36/49 ). The probability that the 5th card is a heart is 13/48. But this needs to be multipled to the rest to get the probability being asked for. So the answer I get for A) is (39/52)* (38/51)* (37/50)*(36/49)* (13/48 ) .

Answer 3

A)
P(exactly 5) = (39/52)(38/51)(37/50)(36/49)(13/48) =
13*39!*47!/(35!*52!) = 0.0822839

B)
P(1) = 0.25
P(2) = 0.19118 = 0.25*39/51 (P(2 and not 1))
P(3) = 0.14529 = 0.19118*38/50
P(4) = 0.10971 = 0.14529*37/49
P(5) = 0.08228 = 0.10971*36/48
p(<6) = 0.77847

C)
P(3) = 0.14529
P(3|<6) = 13/48 = 0.27083 (since the first 2 trials were known to be unsuccessful)

Answer 4

A) What is the probability that exactly 5 cards are required?

prob = (39/52)(38/51)(37/50)(36/49)(13/48) ≈ 0.082

Answer 5

you will need to shuffle a deck seven times to achieve random certainty, so certainly there is a small chance that would happen that the first card delt would begin to appear on the very first shuffle.

small chance would equal 1 in four. you would have 2^n dependent possibilities.