# Deck of Cards Probability: One card missing(randomly), Probability it's a jack?

The question is:

A card is missing from a deck of 52 cards. You draw a card, What's the probability that it's a jack?

I been struggling with probability and would need help understanding this question. What I got so far is that I need to consider two scenerios.

A: Jack is one of the missing cards.
B: Jack is not one of the missing cards.

P(A) = 4/52
P(B) = 48/52

To continue I feel like I might be on the right direction but I could be getting my probabilities wrong and that's what is confusing me and I don't know what do after. Any suggestion are welcomed. Thank you.

• So far you're correct. Now you just have to figure the probability that the card is a Jack in each of the two cases, and apply the law of total probability. Feb 23 at 1:45

Use the events, $$A:$$ the missing card is a Jack, $$B:$$ the missing card is not a Jack, and $$C:$$ a Jack is drawn.

You have $$\mathsf P(A)=4/52$$ and $$\mathsf P(B)=48/52$$

The Law of Total Probability says: $$\mathsf P(C)=\mathsf P(A)\,\mathsf P(C\mid A)+\mathsf P(B)\,\mathsf P(C\mid B)$$

$$\mathsf P(C\mid A)$$ is the probability for drawing a Jack when three are left among the fifty-one remaining cards.

$$\mathsf P(C\mid B)$$ is the probability for drawing a Jack when all four are among the fifty-one remaining cards.

$$~\\~\\~$$

What do you notice about the result?   Is this surprising?

You are correct so far. There are two cases to consider and they are disjoint: the missing card is or is not a jack. If the jack is missing, then there are only $$3$$ possible jacks to draw from the remaining $$51$$-card deck. Similarly, if the missing card is not a jack, then there are $$4$$ possible jacks to draw from the remaining $$51$$-card deck.

Remember how conditional probability works: $$P(A \cap B) = P(B) \, P(A \mid B)$$.

Try to calculate this yourself, but click below to reveal a spoiler when you're ready.

\begin{align*}\def\j{\text{draw jack}}\def\jm{\text{jack missing}}\def\njm{\text{jack not missing}}P(\j) &= P(\j \cap \jm) + P(\j \cap \njm) \\&= P(\jm) \, P(\j \mid \jm) \\&\qquad\qquad {}+ P(\njm) \, P(\j \mid \njm) \\&= \frac{4}{52}\frac{3}{51} + \frac{48}{52}\frac{4}{51} \\&= \frac{1}{13}\frac{3}{51} + \frac{12}{13}\frac{4}{51} \\&= \frac{51}{13 \cdot 51} \\&= \frac{1}{13} \end{align*}

• I got what you got, I remember about conditional probability and the Law of Total Probability. Thank you for the tips. Feb 23 at 2:32