Statistics and probability
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 people simulate this and flip a coin twice. On average, people would get heads on the first flip, and then of them would get heads again. So out of the original people — or of them — would get heads twice in a row.
The number of people we start with doesn't really matter. Theoretically, of the original group will get heads, and 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:
When two events are independent, we can say that
Be careful! This formula only applies to independent events.
Practice problem 1: Rolling dice
Suppose that we are going to roll two fair -sided dice.
Find the probability that both dice show a .
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 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 cards are black, so the probability that the first card is black is . 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 .
Here's what the probabilities would look like in a tree diagram:
So the probability that both cards are black is:
Practice problem 2: Picking students
A table of students has seniors and juniors. The teacher is going to pick students at random from this group to present homework solutions.
Find the probability that both students selected are juniors.
The general multiplication rule
For any two events, we can say that
The vertical bar in 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, .
Want to join the conversation?
- How does the general multiplication rule apply to three events?(38 votes)
- Good question!
For three events A, B, and C, the extension of the general multiplication rule is
P(A and B and C) = P(A)P(B given A)P(C given (A and B)).(76 votes)
- How would you expand P(EFG|H) using the multiplication rule?(5 votes)
- It depends on what you mean by "EFG".
It could be:
- P(HHH|unfair coin)
- P(3,4,5|fair die) [3 or 4 or 5]
- ...(2 votes)
- 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.(4 votes)
- 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 ?(1 vote)
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.(6 votes)
- 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?(3 votes)
- 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.(2 votes)
- if a coin is flipped thrice what is the probability of obtaining atmost 2 tails . please explain by using the above theorems if possible(2 votes)
- "at most" problems are similar to "at least" problems. Instead of calculating all of the possible desired outcomes, we find the complement (undesired) and 'flip' the results to what we actually desire. Let me first explain why finding the desired is too time consuming - "at most 2 tails" means we would have to find the probability for 0 tails, 1 tail, or 2 tails and add all of them together. Instead, finding NOT "at most" only has one possibility, 3 tails. Since we know all of the possibilities together have to be 1, we can take 1 - P(3 tails)=P(at most 2 tails).
With that in mind, here's the actual process: P(3 tails)= P(1 tail)*P(1 tail)*P(1 tail) or (1/2)^3 = 1/8. Since that is the complement of what we desire, we take 1-1/8 to find our actual answer of 7/8. (You can also use a tree diagram for this problem to illustrate this, however it's a good opportunity to use the concept of complements to complete more complicated problems)(2 votes)