What about without replacement Cards are drawn at random and with replacement from an ordinary deck of 52 cards until a spade appears. what is the probability that at least 4 draws are necessary. Is this idea correct. The probability of picking a all non spades on 1 consecutive draw with replacement is 1/4.  2 consecutive is $(1/2)^2$  and 3 consecutive draw is $ (3/4)^3$. So $ 1 - (3/4)^3$ is the final answer. Is it okay to reason that way? What about without replacement?
 A: Let's say you don't replace the cards. There are $52$ cards of which $13$ are spades. You take out a non-spade; this happens with probability $\dfrac{39}{52}=\dfrac{1}{4}$. There are now $51$ cards of which $13$ are spades. You take out a non-spade with probability $\dfrac{38}{51}$. There are now $50$ cards, and so on.
So the probability of it requiring $\ge 4$ picks is
$$\dfrac{39}{52} \cdot \dfrac{38}{51} \cdot \dfrac{37}{50}$$
Because from there, as soon as you've pulled out a spade, at least $4$ picks have happened. The beauty of not replacing the cards is that you have to pull out a spade at some point, so we don't need to worry about the number of picks ever tailing off to $\infty$.
A: Note that the probability of obtaining a spade is 1/4. So for at least four draws, the cards drawn in the first three draws must be a non-spade. Its probability is 3/4. Now for three draws it is (3/4)^3. So for the fourth draw it is a spade. Its probability is 1/4. So my answer is coming as ((3/4)^3).And for with replacement it is coming (39/52)(38/51)(37/50). This is because a spade is not required to come up in the fourth draw as well like clive stated.
A: No, the answer with replacement is $\left(\frac34\right)^4$: $1-\left(\frac34\right)^4$ is the probability that you get at least one space in the first four draws.
Without replacement you can use the same general kind of argument, but the counting is a little different. On the first draw the probability of not getting a spade is $\frac{39}{52}$. If you failed to get a spade on the first draw, there are still $13$ spades left in the deck, but only $51$ cards, so the probability of not getting a spade is $\frac{38}{51}$. Similarly, if neither of the first two cards is a spade, the probability of not getting a spade on the third draw is $\frac{37}{50}$: there are $37$ non-spades left out of the $50$ remaining cards. Finally, if you’ve failed to draw a spade in the first three cards, there are $36$ non-spades left among the remaining $49$ cards, so the probability of getting one of them is $\frac{36}{49}$. Thus, the probability of getting a non-spade on all four draws is
$$\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{37}{50}\cdot\frac{36}{49}\;,\tag{1}$$
and the probability of getting at least one spade in the first four draws is
$$1-\frac{39\cdot38\cdot37\cdot36}{52\cdot51\cdot50\cdot49}\;.$$
The difference from the probability with replacement is fairly small: $\left(\frac34\right)^4=0.31640625$, and the product $(1)$ is about $0.30381753$. The more you draw, the more the two probabilities will diverge: after $40$ draws the probability of having drawn no spades is obviously $0$ without replacement, but with replacement it’s a little more than $10^{-5}$. That’s small, but certainly not $0$.
A: Probability of not getting a spade is $3/4$. For this to happen three times in a row, the probability is $(3/4)^3$.
If you want without replacement, the second time and the third time will give you different probabilities. The second time, you have lost $1$ non-spade card, so the probability of not getting a spade becomes $38/51$. The third time it is $37/50$. So the final answer is $(39\cdot 38\cdot 37)/(52\cdot 51\cdot 50)$.
A: I derived the general formula for the probability of drawing at least one of a sected type of card from a deck in a certain number of draws, where
t = total number of cards in deck
n = number of card type to select (i.e. spade=13)
c = number of cards drawn
P = probability you will draw at least 1 of selected type
such that
P=1-((t-n)!(t-c)!)/(t!(t-c-n)!)
so.... for your problem: t=52 cards, n=13 spades, c=4 cards
we have
P = 1-((52-13)!(52-4)!)/(52!(52-13-4)!)
P = 1-(39!48!)/(52!35!)
P = 14498/20825
P = 0.696182472989...
P = 69.618... % chance of drawing at least 1 spade within 4 cards
This is the same as Brian's answer, but this does not completely answer your question.
You need to somehow take this figure and then reverse it... IDK, hopefully someone else can elaborate. GOOD LUCK!
