The General Multiplication Rule (Explanation & Examples)


The general multiplication rule states that the probability of any two events, A and B, both happening can be calculated as:

P(A and B) = P(A) * P(B|A)

The vertical bar | means “given.” Thus, P(B|A) can be read as “the probability that B occurs, given that A has occurred.”

If events A and B are independent, then P(B|A) is simply equal to P(B) and the rule can be simplified to:

P(A and B) = P(A) * P(B)

Let’s walk through a few examples of both independent and dependent events to see how we can apply this general multiplication rule in practice.

The General Multiplication Rule for Dependent Events

The following examples illustrate how to use the general multiplication rule to find probabilities related to two dependent events. In each example, the probability that the second event occurs is affected by the outcome of the first event.

Example 1: Balls in an Urn

An urn contains 4 red balls and 3 green balls. Bob is going to randomly select 2 balls from the urn, without replacement. What is the probability that he chooses 2 red balls?

Solution: The probability that he selects a red ball on the first attempt is 4/7. Once that ball is then removed, the probability that he selects a red ball on the second attempt is 3/6. Thus, the probability that he selects 2 red balls can be calculated as:

P(Both red) = 4/7 * 3/7 ≈ 0.2249

Example 2: Cards in a Deck

A deck of cards contains 26 black cards and 26 red cards. Debbie is going to randomly select 2 cards from the deck, without replacement. What is the probability that she chooses 2 red cards?

Solution: The probability that she selects a red card on the first attempt is 26/52. Once that card is then removed, the probability that she selects a red card on the second attempt is 25/51. Thus, the probability that she selects 2 red cards can be calculated as:

P(Both red) = 26/52 * 25/51 ≈ 0.2451

The General Multiplication Rule for Independent Events

The following examples illustrate how to use the general multiplication rule to find probabilities related to two independent events. In each example, the probability that the second event occurs is not affected by the outcome of the first event.

Example 1: Flipping Two Coins

Suppose we flip two coins. What is the probability that both coins land on heads?

Solution: The probability that the first coin lands on heads is 1/2. No matter which side the first coin lands on, the probability that the second coin lands on heads is also 1/2. Thus, the probability that both coins land on heads can be calculated as:

P(Both land on heads) = 1/2 * 1/2 = 0.25

Example 2: Rolling Two Dice

Suppose we roll two dice at once. What is the probability that both dice land on the number 1?

Solution: The probability that the first dice lands on “1” is 1/6. No matter which side the first dice lands on, the probability that the second dice lands on “1” is also 1/6. Thus, the probability that both dice land on “1” can be calculated as:

P(Both land on “1”) = 1/6 * 1/6 = 1/36 ≈ 0.0278

8 Replies to “The General Multiplication Rule (Explanation & Examples)”

  1. Shouldn’t this
    P(Both red) = 4/7 * 3/7 ≈ 0.2249 be P(Both red) = 4/7 * 3/6 ≈ 0.2857?
    You have already picked the red ball and are not replacing it?

  2. Hi Zach,
    In one of the examples mentioned by you (below):

    Example 1: Balls in an Urn
    P(Both red) = 4/7 * 3/7 ≈ 0.2249
    Example 2: Cards in a Deck
    P(Both red) = 26/52 * 25/51 ≈ 0.2451

    The denominator rule for both is different. Is there a mistake? According to me in example 1, the probability should be calculated as 4/7 * 3/6.

  3. Hello Zach,

    Thanks a lot for your website. It’s been very helpful.

    I’m sorry to reply only because of a mistake, but I wanted to let you know about a small typo.

    at Example 1: Balls in an urn, in the final formula it should be 3/6, instead of 3/7, right?

    Thank you!

  4. In above Example 1: Balls in an Urn
    if a red ball on the second attempt is 3/6, then wouldn’t
    P(Both red) = 4/7 * 3/6 ≈ 0.285 ?

  5. Hi Zach,

    Thank you for making stats approachable. I wish I had found your site sooner.

    I think that your first example has an error in it. The final sum reads “P(Both red) = 4/7 * 3/7 ≈ 0.2249”, but I think it should read “ P(Both red) = 4/7 * 3/6≈ 0.2857” given that the events are dependent.

  6. Solution: The probability that he selects a red ball on the first attempt is 4/7. Once that ball is then removed, the probability that he selects a red ball on the second attempt is 3/6. Thus, the probability that he selects 2 red balls can be calculated as:

    P(Both red) = 4/7 * 3/7 ≈ 0.2249 shouldnt it be like 3/6 instead of 3/7

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