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## Statistics and probability

### Course: Statistics and probability>Unit 7

Lesson 9: Conditional probability and independence

# Conditional probability and independence

Use conditional probability to see if events are independent or not.

## Want to join the conversation?

• How does the logic work when we conclude that P(delayed) is independent to P(delayed | snowy) when they are approximately the same?
• When they are approx the same, it means that the probability of a delay is the same whether or not it snows. In other words, the probability of delay has nothing to do with whether or not it snows; the event Delay is independent from the event Snow. Not sure if that helps/answers your question! :)
• How do we quantify or truly understand what it means to be "so much higher"?
Thanks!
• how do you do this in Venn diagram form?
• why do we do (delayed|snowy) but not (snowy|delayed)? How do we know which one to use for future questions that are similar.
• I do not understand how can we understand that is it independent or dependent the problem after solving it.
(1 vote)
• Intuitively, in other case whenever the value of P(delayed) = P(delayed | snowy), we could see that with or without the snowy condition, the probability of the flight delay stays the same. So we can conclude that they're independent to each other. But in this case, their values are different. So the existence snowy condition affects the flight delay probability and they're dependent.
• I don't understand how to find if it is independent or dependent. What makes something independent or dependent?
• When something is independent, then the probability of A given B is the same as the probability of just A. In other words, both the probability of A and probability of B do not affect one another.
• At ,Sal talks about the existence of a theoretical probability value. I do not quite understand this. Does this mean that there is an exact real number assigned to every probability we try to calculate and even though it is magically hidden from us, it somehow exists in some place?(Where,actually?) I understand that the more sample we have, the more confident we can be about our calculated probability but this notion of absolute theoretical probability is bugging me.
• I have this problem in my book and I am very frustrated.
Four trumpet players' instruments are mixed up and the trumpets are given to the players just before concert. What is the probability that no one gets his or her trumpet back. The answer in the book is 3/8
• Interesting problem!

We can use the inclusion-exclusion principle to find the probability that at least one player gets his/her trumpet back. Then we can subtract this from 1.

Let A, B, C, D represent the events that the first, second, third, and fourth player, respectively, individually gets his/her trumpet back.

Note that the probability that a specific collection of k of these events all occur is 1/(4Pk).

The probability that at least one player gets his/her trumpet back is

P(AUBUCUD)

= [P(A) + P(B) + P(C) + P(D)] - [P(A n B) + P(A n C) + P(A n D) + P(B n C) + P(B n D) + P(C n D)] + [P(A n B n C) + P(A n B n D) + P(A n C n D) + P(B n C n D)] - P(A n B n C n D)

= (4C1)/(4P1) - (4C2)/(4P2) + (4C3)/(4P3) - (4C4)/(4P4)

= 1/1! - 1/2! + 1/3! - 1/4!
= 1 - 1/2 + 1/6 - 1/24
= 5/8.

So the probability that no player gets his/her trumpet back is 1-5/8 = 3/8.

We can extend this to n players. A similar calculation gives a final answer of
1/2! - 1/3! + 1/4! - ... + (-1)^n / n!.
This implies that as n becomes large, the final answer approaches 1/e.

Have a blessed, wonderful day!
• how do we know, wen should we use additive law and multiplicative and conditional probability.
(1 vote)
• Roughly, “or” means “add”; “and” means multiply, but sometimes with modifications.

If A and B are mutually exclusive (that is, they can’t both occur), then
P(A or B) = P(A) + P(B).

Without the assumption of mutual exclusivity, we have to modify this rule by subtracting the probability of overlap:
P(A or B) = P(A)+P(B)-P(A and B).

If A and B are independent (that is, the occurrence of a specific one of these two events does not influence the probability of the other event), then
P(A and B) = P(A)P(B).

Without the assumption of independence, we have to modify this rule by replacing one of the individual probabilities by a conditional probability:

P(A and B) = P(A given B)P(B) assuming P(B) is nonzero;

P(A and B) = P(A)P(B given A) assuming P(A) is nonzero.

Have a blessed, wonderful day!