Probability A Intersection B Formula with Examples

Key Points

• The probability a intersection b formula is $P(A \cap B) = P(A) \times P(B \mid A)$, the general multiplication rule.

• Multiplying $P(A)$ by $P(B)$ is valid only when the two events are independent, so that rule is a special case rather than the main one.

• Rearranged, the same relationship gives conditional probability as $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.

• The intersection can also be recovered from the union through $P(A \cap B) = P(A) + P(B) - P(A \cup B)$.

• Mutually exclusive events have $P(A \cap B) = 0$, which makes them the opposite case from independent events rather than the same thing.

What Is the Probability A Intersection B Formula?

The probability a intersection b formula gives the chance that two events both happen:

$$P(A \cap B) = P(A) \times P(B \mid A)$$

When the events are independent, $P(B \mid A) = P(B)$ and the rule simplifies:

$$P(A \cap B) = P(A) \times P(B) \qquad \text{(independent events only)}$$

This article treats the probability sense. The set-theory operation of the same name, intersecting two collections of elements, is covered by intersection of sets.

Why the Conditional Term Is Needed

Asking for both events means asking for $A$ to happen and then for $B$ to happen given that $A$ already has. The second step is conditional by construction.

Drawing two cards makes this concrete. The chance the first is an ace is $\frac{4}{52}$, but the chance the second is also an ace is $\frac{3}{51}$, because one ace has gone and one card has left the deck.

Independence is the situation where that adjustment is unnecessary. Rolling a die twice leaves the second roll untouched by the first, so $P(B \mid A) = P(B)$ and the rule collapses to a plain product.

Variable Key

Examples of the Probability A Intersection B Formula

Example 1

A fair die is rolled twice. Find the probability of a six on both rolls.

The rolls are independent, so the conditional term equals $P(B)$:

$$P(A \cap B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$$

Final answer: $\frac{1}{36}$

Example 2

Two cards are drawn without replacement. Find the probability both are aces, by multiplying the two probabilities.

Treating the draws as independent gives

$$\frac{4}{52} \times \frac{4}{52} = \frac{1}{169}$$

But the first card is not replaced, so by the second draw only 3 aces remain among 51 cards. The second probability is $\frac{3}{51}$, not $\frac{4}{52}$.

Using the general rule:

$$P(A \cap B) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$

Final answer: $\frac{1}{221}$. Multiplying unconditional probabilities overstates the chance whenever the draws are dependent.

Example 3

$P(A) = 0.6$ and $P(B \mid A) = 0.5$. Find $P(A \cap B)$.

$$P(A \cap B) = 0.6 \times 0.5 = 0.3$$

Final answer: 0.3

Example 4

$P(A) = 0.7$, $P(B) = 0.5$ and $P(A \cup B) = 0.9$. Find $P(A \cap B)$ and say whether the events are independent.

Rearranging the union rule:

$$P(A \cap B) = 0.7 + 0.5 - 0.9 = 0.3$$

For independence the product would have to be $0.7 \times 0.5 = 0.35$.

Final answer: $P(A \cap B) = 0.3$, and since $0.3 \neq 0.35$ the events are not independent.

Comparing the actual intersection against the product is the standard test for independence.

Example 5

A bag holds 5 red and 3 blue balls. Two are drawn without replacement. Find the probability both are red.

$$P(A \cap B) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$

Final answer: $\frac{5}{14}$

The second fraction has both a smaller numerator and a smaller denominator, since one red ball and one ball in total have gone.

Independent, Dependent and Mutually Exclusive

Independent and mutually exclusive are routinely confused, and they are near opposites. Independent events can happen together and simply do not influence each other, while mutually exclusive events cannot happen together at all.

Where Do Students Trip Up on P(A and B)?

Mistake 1: Multiplying without checking independence

Where it slips in: The simplified rule $P(A)P(B)$ is the version usually memorised, and it works on dice and coins.

Don't do this: $\frac{4}{52} \times \frac{4}{52}$ for two cards drawn without replacement.

The correct way: Use $P(A)P(B \mid A)$, adjusting the second probability. Ask whether the first event changes the situation for the second before multiplying anything.

Students who look for the phrase "without replacement" in the question before writing anything get this right consistently, because those two words are what make the events dependent. Their absence is equally informative.

Mistake 2: Confusing independent with mutually exclusive

Where it slips in: Both terms describe a relationship between two events and both sound like a kind of separateness.

Don't do this: Set $P(A \cap B) = 0$ because the events are independent.

The correct way: Independent events have $P(A \cap B) = P(A)P(B)$, which is zero only if one of them is impossible. Mutually exclusive events are the ones with a zero intersection.

Mistake 3: Letting the intersection exceed one of the events

Where it slips in: The arithmetic is done without a sanity check on the result.

Don't do this: Report $P(A \cap B) = 0.8$ when $P(A) = 0.6$.

The correct way: The intersection can never exceed either individual probability, since both events happening is a stricter requirement than either alone. That bound catches most arithmetic slips immediately.

Key Takeaways

• The probability a intersection b formula is $P(A \cap B) = P(A)P(B \mid A)$.

• The simpler product $P(A)P(B)$ applies only to independent events, so independence must be checked rather than assumed.

• Rearranging gives both conditional probability and the intersection from the union.

• Independent and mutually exclusive are near opposites, not synonyms.

• To work through probability with a teacher, explore a high school math tutor or math classes online.

Practice These Three Before Moving On

A bag holds 6 green and 4 yellow counters. Find the probability of drawing two greens without replacement, then with replacement. Finally, given $P(A)=0.5$, $P(B)=0.4$ and $P(A \cup B)=0.7$, decide whether $A$ and $B$ are independent.

Want a live Bhanzu trainer to walk through more probability problems? Book a free demo class.

Read More

• Conditional Probability Formula

• What Is Probability

• Operations on Sets

• Venn Diagram

• Union of Sets