# If three cards are drawn from a standard deck, what is the probability of drawing a $10, J, Q$ in that order with replacement? without replacement?

If I get to draw $3$ cards from a standard deck.

With replacement - a) What is the probability of drawing a $10,J,Q$ in that order?

Without replacement - b) What is the probability of drawing a $10,J,Q$ in that order?

With replacement - c) Drawing at least one ace (I got $= \left(\frac{4}{52}\right)^3$).

Without replacement - d) Drawing at least one ace (I got $= \left(\frac{4}{52}\right)\left(\frac{3}{51}\right)\left(\frac{2}{50}\right)$)

Having trouble with a) and b) I know that order matters so I would have to use combinations.

Any thoughts? Thanks.

a) In other words, it must be the case that you get a 10, then a J, then a Q. Since the draws are with replacement, then the probabilities don't change and the draws are independent from one another (assuming the deck is shuffled after replacement). Then we have $$P(TJQ) = P(T)P(J)P(Q) = \frac{4}{52}\cdot\frac{4}{52}\cdot\frac{4}{52}.$$

b) Again, it must be the case that you get a 10, then a J, then a Q. Since the draws are without replacement, then the probabilities do change and the draws are not independent from one another. After each draw, the number of cards in deck reduces by one, and hence $$P(TJQ) = \frac{4}{52}\cdot \frac{4}{51}\cdot\frac{4}{50}.$$

c) No, what you calculated is the chance that you got exactly 3 Aces (or really any rank) in three draws with replacement.

A direct calculation involves find the chance of an Ace in the first or second or third draw and using inclusion exclusion.

Alternatively, we can use the complement: No Aces in three draws. This is equal to $(48/52)^3$. Hence $$P(\text{At least one Ace}) = 1-P(\text{No Aces}) = 1-\left(\frac{48}{52}\right)^3.$$

d) Again, we use the complement. There are $48$ Non Aces, and in three draws we must draw a non Ace. After each draw, the number of non Aces goes down, and so $$P(\text{At least one Aces})=1- P(\text{No Aces}) = 1-\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}.$$

Some useful hints:

Ad a) The probability that the first card is a 10 is $\frac{4}{52}=\frac{1}{13}$. You have four 10´s and 52 cards in total. Similar probabilities for $J$ and $Q$.

Ad b) The probability that the first card is a 10 is $\frac{4}{52}=\frac{1}{13}$. But then the probability that the next card is a J is $\frac{4}{52-1}$, because the remaining cards have been reduced by one card.

Ad c) Here you can calculate by using the converse probability. The probability to draw at least on ace is equal to One minus the probability to draw no ace: $P(\texttt{"at least one ace"})=1-P(\texttt{"no ace"})$

Ad d) Same idea like in c), but without replacement. $P(\texttt{"First card no ace")}=\frac{48}{52}$, $P(\texttt{"Second card no ace")}=\frac{48-1}{52-1}=\frac{47}{51}$ and so on.