From a Conditional Probability to a Joint Probability
In Conditional Probability with Tree Diagrams, we multiplied branch probabilities along a path to find the probability of a complete sequence of results. The general multiplication rule expresses that same idea for two events: the probability that both occur is the probability of the first event multiplied by the probability of the second event given that the first has occurred.
This rule is especially useful for drawing cards without replacement. Once the first card is drawn, it is no longer in the deck. The composition of the deck changes, so the probability of a particular second draw may depend on the first card. The conditional probability in the rule accounts for that change.
The condition \(P(A)>0\) matters because \(P(B\mid A)\) is defined only when \(P(A)>0\). The events can be dependent or independent; the general rule applies either way. If \(A\) and \(B\) are independent, the conditional probability \(P(B\mid A)\) equals \(P(B)\), giving the independence multiplication rule covered in Mutually Exclusive Versus Independent Events.
The events in an intersection can also be written in the opposite order. If \(P(B)>0\), then \(P(A\cap B)=P(B)P(A\mid B)\). This reversed form requires \(P(B)>0\), because that condition is what makes \(P(A\mid B)\) defined. Choose the order that makes the conditional probability easiest to find.
Using the Rule for Two Card Draws
Assume a standard deck has 52 cards, with 4 suits and 13 cards in each suit. When two cards are drawn one after the other without replacement, there are 52 possible cards for the first draw and then 51 cards remaining for the second. But the number of cards that meet the second event’s description depends on what happened on the first draw.
For example, if the first card is an ace, only 3 aces remain among the 51 cards. Thus, the probability that the second card is an ace, given that the first card was an ace, is \(3/51\). The unconditional probability that a randomly selected card is an ace, \(4/52\), is not the right second factor for this sequence.
This is the same path multiplication used in the earlier tutorial on tree diagrams. A complete path has one branch for each draw. Multiplying the branch probabilities gives the probability of that path. If several different paths satisfy the event in the question, find each path probability and add them, as in the earlier tutorials on the general addition rule and disjoint events.
Worked Example: Drawing Two Aces
Worked Example: Drawing Two Aces
Two cards are drawn from a standard deck without replacement. Find the probability that both cards are aces.
State: Let \(A\) mean that the first card is an ace, and let \(B\) mean that the second card is an ace. We want \(P(A\cap B)\), the probability that both draws are aces.
Plan: Use the general multiplication rule. The first draw has 4 aces among 52 cards. Given that the first card was an ace, 3 aces remain among 51 cards for the second draw. Since \(P(A)=4/52>0\), the conditional probability \(P(B\mid A)\) is defined.
Do: Substitute the first-draw probability and the conditional second-draw probability into the rule:
The fraction reduces because both 12 and 2652 are divisible by 12. As a check, there are \(4\cdot3=12\) ordered ways to draw two different aces, out of \(52\cdot51=2652\) ordered pairs of cards. This gives the same fraction, \(12/2652=1/221\).
Conclude: The probability of drawing two aces in a row without replacement is \(1/221\), or about 0.0045. The second factor is \(3/51\), not \(4/52\), because one ace has already been removed.
The Second Probability Depends on the First Result
A common question is whether the two probabilities in a multiplication should always be the same. They should not. The general multiplication rule calls for \(P(B\mid A)\), which is specifically the probability of the second event given the first. In a no-replacement draw, this conditional probability often differs from the probability of the second event before any card is drawn.
The change need not always be large, and it depends on how the events are defined. For example, after drawing a red card, the number of black cards still in the deck is unchanged, even though there is one fewer card in total. That gives a second-draw probability of \(26/51\) for black, rather than the original \(26/52\). The next example uses this distinction.
Worked Example: Red First, Then Black
Worked Example: Red First, Then Black
Two cards are drawn without replacement from a standard deck. Find the probability that the first card is red and the second card is black.
State: Let \(R\) mean that the first card is red, and let \(K\) mean that the second card is black. We want \(P(R\cap K)\).
Plan: Apply \(P(R\cap K)=P(R)P(K\mid R)\). There are 26 red cards among 52 cards initially. Once a red card is drawn, all 26 black cards remain, but only 51 cards remain altogether. The condition \(P(R)>0\) holds.
Do: Calculate the two factors and multiply:
Check the arithmetic using the un-reduced product: \(26\cdot26=676\) and \(52\cdot51=2652\), so the product is \(676/2652\). Dividing numerator and denominator by 52 gives \(13/51\), as above.
Conclude: The probability that the first card is red and the second is black is \(13/51\), or about 0.2549. It is slightly greater than \(1/4\), the result of multiplying \(26/52\) by \(26/52\), because after the red card is removed, the 26 black cards make up a slightly larger share of the 51 remaining cards.
When Either Draw Order Works
Some questions describe what the two cards are, without requiring a particular order. For example, “one king and one queen” includes a king followed by a queen and a queen followed by a king. Those are two distinct, non-overlapping paths: a particular ordered pair cannot have both orders at once.
Use the multiplication rule to find the probability of each order separately. Then add the path probabilities because either order satisfies the event. This combines the general multiplication rule for each sequence with the addition rule for the complete event.
Worked Example: One King and One Queen in Either Order
Worked Example: One King and One Queen in Either Order
Two cards are drawn without replacement from a standard deck. Find the probability that one card is a king and the other is a queen, in either order.
State: Let \(KQ\) be the event that the first card is a king and the second is a queen. Let \(QK\) be the event that the first card is a queen and the second is a king. The event in the question is \(KQ\cup QK\).
Plan: Find \(P(KQ)\) and \(P(QK)\) using the multiplication rule. The two paths are disjoint, so add their probabilities. For either order, there are 4 cards of the required rank on the first draw and 4 of the other rank among the 51 cards remaining.
Do: For king first and queen second:
For queen first and king second, the counts are the same, so:
The two paths cannot occur simultaneously, so add their probabilities. The sum is \(8/663\), which is approximately 0.0121. Checking before reduction, \(16/2652+16/2652=32/2652\); dividing top and bottom by 4 gives \(8/663\).
Conclude: The probability of drawing one king and one queen in either order is \(8/663\), or about 0.0121. Counting only king then queen would omit the queen-then-king path and give half the desired probability.
A Reliable Routine for the General Multiplication Rule
State what \(A\) and \(B\) mean, including which event refers to the first draw and which to the second.
Use \(P(A\cap B)=P(A)P(B\mid A)\) when \(P(A)>0\). If using the reversed form, verify that \(P(B)>0\).
Use the cards available before the first draw. For a standard deck, there are 52 cards at the start.
For the conditional factor, count the cards that meet the second event among the cards remaining after the first event.
Multiply the factors for one path. If several disjoint paths meet the event, add their probabilities and interpret the final result in context.
Common Mistakes and AP Exam Tips
- Using the unconditional probability for the second draw. With no replacement, the second-draw probability must reflect the first result. A full-credit solution identifies the remaining number of cards in both the numerator and denominator.
- Multiplying the marginal probabilities automatically. \(P(A)P(B)\) is the independence rule, not the general rule for every pair of events. For the general rule, use \(P(A)P(B\mid A)\). In the red-then-black example, the correct second factor is \(26/51\), not \(26/52\).
- Forgetting the condition required by the formula. The expression \(P(A)P(B\mid A)\) requires \(P(A)>0\). If you reverse the events and write \(P(B)P(A\mid B)\), state or verify that \(P(B)>0\) instead.
- Adding probabilities for the stages of one path. A path requires the first result and then the second result, so multiply. Add only when there are distinct paths that each satisfy the event, as with king-queen and queen-king.
- Leaving out an allowed order. If the question does not specify which rank comes first, list the possible orders before calculating. “One king and one queen” includes both orders.
- Giving a number without identifying the event. A strong conclusion names the requested outcome, such as “the probability that the first card is red and the second is black is about 0.2549.” This makes clear that the answer is for the requested sequence.
A quick check is to ask what the conditional probability means in words. For example, \(P(K\mid R)\) means the probability that the second card is black among draws where the first card was red. If the denominator in your fraction does not describe the cards available within that condition, revisit the setup.
Check Your Understanding
For each question, define the events, show the conditional probability, and give the result in context.
- Two cards are drawn without replacement. Find the probability that both are jacks.
- Two cards are drawn without replacement. Find the probability that the first card is black and the second is red.
- Two cards are drawn without replacement. Find the probability that the first is a heart and the second is the ace of hearts.
- Two cards are drawn without replacement. Find the probability of drawing one ten and one nine, in either order.
- In a sentence, explain why \(P(A\cap B)\) can be found using \(P(B)P(A\mid B)\) only when \(P(B)>0\).