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The general multiplication rule

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When we calculate probabilities involving one event AND another event occurring, we multiply their probabilities.
In some cases, the first event happening impacts the probability of the second event. We call these dependent events.
In other cases, the first event happening does not impact the probability of the seconds. We call these independent events.

Independent events: Flipping a coin twice

What is the probability of flipping a fair coin and getting "heads" twice in a row? That is, what is the probability of getting heads on the first flip AND heads on the second flip?
Imagine we had 100 people simulate this and flip a coin twice. On average, 50 people would get heads on the first flip, and then 25 of them would get heads again. So 25 out of the original 100 people — or 1, slash, 4 of them — would get heads twice in a row.
The number of people we start with doesn't really matter. Theoretically, 1, slash, 2 of the original group will get heads, and 1, slash, 2 of that group will get heads again. To find a fraction of a fraction, we multiply.
We can represent this concept with a tree diagram like the one shown below.
We multiply the probabilities along the branches to find the overall probability of one event AND the next even occurring.
For example, the probability of getting two "tails" in a row would be:
P, left parenthesis, start text, T, space, a, n, d, space, T, end text, right parenthesis, equals, start fraction, 1, divided by, 2, end fraction, dot, start fraction, 1, divided by, 2, end fraction, equals, start fraction, 1, divided by, 4, end fraction
When two events are independent, we can say that
P, left parenthesis, start text, A, space, a, n, d, space, B, end text, right parenthesis, equals, P, left parenthesis, start text, A, end text, right parenthesis, dot, P, left parenthesis, start text, B, end text, right parenthesis
Be careful! This formula only applies to independent events.

Practice problem 1: Rolling dice

Suppose that we are going to roll two fair 6-sided dice.
problem 1
Find the probability that both dice show a 3.
Choose 1 answer:
Choose 1 answer:

Dependent events: Drawing cards

We can use a similar strategy even when we are dealing with dependent events.
Consider drawing two cards, without replacement, from a standard deck of 52 cards. That means we are drawing the first card, leaving it out, and then drawing the second card.
What is the probability that both cards selected are black?
Half of the 52 cards are black, so the probability that the first card is black is 26, slash, 52. But the probability of getting a black card changes on the next draw, since the number of black cards and the total number of cards have both been decreased by 1.
Here's what the probabilities would look like in a tree diagram:
So the probability that both cards are black is:
P, left parenthesis, start text, b, o, t, h, space, b, l, a, c, k, end text, right parenthesis, equals, start fraction, 26, divided by, 52, end fraction, dot, start fraction, 25, divided by, 51, end fraction, approximately equals, 0, point, 245

Practice problem 2: Picking students

A table of 5 students has 3 seniors and 2 juniors. The teacher is going to pick 2 students at random from this group to present homework solutions.
problem 2
Find the probability that both students selected are juniors.
Choose 1 answer:
Choose 1 answer:

The general multiplication rule

For any two events, we can say that
P, left parenthesis, start text, A, space, a, n, d, space, B, end text, right parenthesis, equals, P, left parenthesis, start text, A, end text, right parenthesis, dot, P, left parenthesis, start text, B, end text, vertical bar, start text, A, end text, right parenthesis
The vertical bar in P, left parenthesis, start text, B, end text, vertical bar, start text, A, end text, right parenthesis means "given," so this could also be read as "the probability that B occurs given that A has occurred."
This formula says that we can multiply the probabilities of two events, but we need to take the first event into account when considering the probability of the second event.
If the events are independent, one happening doesn't impact the probability of the other, and in that case, P, left parenthesis, start text, B, end text, vertical bar, start text, A, end text, right parenthesis, equals, P, left parenthesis, start text, B, end text, right parenthesis.

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  • piceratops seed style avatar for user Kristin
    How would you expand P(EFG|H) using the multiplication rule?
    (5 votes)
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  • aqualine seed style avatar for user guncha.ch
    does term "given" , means we need to subtract the P(A) from (B). I almost understood the topic just the last formula P(B/A), makes me little uncomfortable.
    (3 votes)
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  • blobby green style avatar for user Steve
    How does the general multiplication rule apply to three events?
    (2 votes)
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  • aqualine seed style avatar for user Maxim Lifirenko
    Can someone explain how this, "A dice is thrown 5 times. If getting an even number is a success, find the probability of all 5 successes."

    Equals = 1/32 ?
    (2 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      Yes.

      There are 6 possible and equally likely outcomes for each die, of which 3 are even numbers.
      So, the probability of rolling an even number on a die is 3∕6 = 1∕2.

      Since the five dice are independent events, we can multiply their probabilities together, so the probability that all five dice show even numbers is (1∕2)⁵ = 1∕32.
      (2 votes)
  • blobby green style avatar for user junejoiqra.6
    How we find the probability of one event out of three events when they are dependent events.
    (2 votes)
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  • blobby green style avatar for user DiMarco
    How do you know which one to multiply
    (2 votes)
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  • starky seed style avatar for user Quinn Ciccoretti
    Is there a proof for the last statement? I've heard it referred to as a definition.
    (1 vote)
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  • blobby green style avatar for user sweta jha
    A coin is tossed repeatedly. the coin is unfair and p(H)= p. the game ends, the first time that two consecutive heads (HH) or two consecutive tails (TT) are observed. the player wins if HH is observed and losses if TT is observed. find the probability that player wins.
    (1 vote)
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  • aqualine ultimate style avatar for user Unai
    No lo entendí muy bien.
    Aun así buen ejercicio.
    Un saludo.
    (1 vote)
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  • leafers sapling style avatar for user Rian
    I've been given an example that says: "A person has a probability of getting pulled over for speeding as 0.8; a person that is pulled over has a probability of getting a ticket as 0.9". This is a P(E and F) = P(E)*P(F|E), which comes out to .72.
    What I am asking, is if the person is pulled over period, why wouldn't they have a .9 chance of getting a ticket.

    I have answered my own question while typing this out. I believe that there is a .72 chance of a person getting a ticket while being pulled over for speeding due to the probablity of even getting pulled over for speeding?
    (1 vote)
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