What is Conditional probability: Definition and 242 Discussions

In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes PB(A) or P(A/B). For example, the probability that any given person has a cough on any given day may be only 5%. But if we know or assume that the person is sick, then they are much more likely to be coughing. For example, the conditional probability that someone unwell is coughing might be 75%, in which case we would have that P(Cough) = 5% and P(Cough|Sick) = 75%.
Conditional probability is one of the most important and fundamental concepts in probability theory. But conditional probabilities can be quite slippery and might require careful interpretation. For example, there need not be a causal relationship between A and B, and they don't have to occur simultaneously.
P(A|B) may or may not be equal to P(A) (the unconditional probability of A). If P(A|B) = P(A), then events A and B are said to be independent: in such a case, knowledge about either event does not alter the likelihood of each other. P(A|B) (the conditional probability of A given B) typically differs from P(B|A). For example, if a person has dengue, they might have a 90% chance of testing positive for dengue. In this case, what is being measured is that if event B ("having dengue") has occurred, the probability of A (test is positive) given that B (having dengue) occurred is 90%: that is, P(A|B) = 90%. Alternatively, if a person tests positive for dengue, they may have only a 15% chance of actually having this rare disease, because the false positive rate for the test may be high. In this case, what is being measured is the probability of the event B (having dengue) given that the event A (test is positive) has occurred: P(B|A) = 15%. Falsely equating the two probabilities can lead to various errors of reasoning such as the base rate fallacy. Conditional probabilities can be reversed using Bayes' theorem.
Conditional probabilities can be displayed in a conditional probability table.

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  1. Hamiltonian

    Finding a conditional probability from joint p.d.f

    using the equation mentioned under Relevant Equations I can get, $$\mathbb{P}(2X > Y |1 < 4Z < 3) = \frac{\mathbb{P}(2X>Y, 1<4z<3)}{\mathbb{P}(1<4z<3)}$$ I can find the denominator by finding the marginal probability distribution, ##f_{Z}(z)## and then integrating that with bounds 0 to 1. But I...
  2. MathMan2022

    Conditional probability problem

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  3. V

    Expected Value of Election Results

    I submitted this solution, and it was marked incorrect. Could I get some feedback on where I went wrong? Let S represent the event that Party A wins the senate and H represent the event that Party A wins the house. There are 4 cases: winning the senate and house (##S \cap H##), winning just...
  4. WMDhamnekar

    MHB Conditional probability

    North and south have ten trumps between them ( trumps being cards of specified suit). (a) Find the probability that all three remaining trumps are in the same hand. (that is either east or west has no trumps). (b) If it is known that king of trumps is included among the three, what is the...
  5. C

    I Taking socks out of drawers, conditional probability

    Problem: In a dresser there are 3 drawers. In one drawer there are two black socks and one white sock, in the second drawer there are two white socks, and in the third drawer there is a black and white sock. Suppose I chose a drawer randomly ( meaning, in a uniform distribution ) and I took a...
  6. S

    Conditional probability of a test records this positive result

    My attempt: $$P(\text{B is positive}|\text{A is positive})=\frac{P(\text{B is positive} \cap \text{A is positive})}{P(\text{A is positive})}$$ $$=\frac{P(\text{B is positive})\times P(\text{A is positive})}{P(\text{A is positive})}$$ $$=P(\text{B is positive})$$ $$=0.01 \times 0.99 + 0.99 \times...
  7. P

    B Decision for conditional probability instead of intersection of events

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  8. Moara

    Conditional probability and criminal DNA analysis

    We know that ##P(A-) = (95\% \cdot 0.5\% + 5\% \cdot 98.5\% )## and ##P(guilty \ and \ A-) = (95\% \cdot 0.5\%)##, so letter a) is just ##P(guilty \ and \ A-)/P(A-)##. What I tried to do in letter b) was again using the conditional probability theorem. First calculating the probability that...
  9. Moara

    Observation of events and analysis of the associated Hypotheses

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  10. chwala

    Solve the conditional probability question

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  11. M

    Probability notation: question about joint and conditional probability

    Hi, Just a quick question about conditional and marginal probabilities notation. Question: What does ## p(a|b, c) ## mean? Does it mean: 1) The probability of A, given (B and C) - i.e. ## p[A | (B \cap C)] ## OR 2) The probability of (A given B) and C - i.e. ## p[(A | B) \cap C] ## I was...
  12. jisbon

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    Confused and not sure if it is correct, but please do correct my steps. We let event B be that there are at least 3 customers entering in 5 minutes. Hence P(B) = 1- P(X=0)- P(X=1) - P(X=2) = ##1- \dfrac{e^{-5}5^{0}}{0!}-\dfrac{e^{-5}5^{1}}{1!}-\dfrac{e^{-5}5^{2}}{2!} ## = 0.8753... Now we let...
  13. Addez123

    Conditional probability of dying from eating a poison fruit

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  14. CaptainX

    B What is Conditional Probability and its Properties?

    1. Definition If E and F are two events associated with the same sample space of a random experment, the conditional probability of the event E given that F has occurred, i.e. P(E|F) is given by P(E|F) = (E∩F)/P(F) (P≠0) 2. Properties of conditional probability Let E and F be events of...
  15. Manasan3010

    Is an answer possible - Conditional Probability

    I am a noob to this topic so correct me If I made any silly mistake. By plugging in the values I managed to get p(abc)=0.75*0.9*p(c|ab) Here How can I find p(c|ab)? Is this question unsolvable or can I derive it? I also want to know what is meant by p(abc) in literary terms. I also created a...
  16. Calculuser

    I A Lottery Game With Conditional Probability?

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  17. Eclair_de_XII

    How do I derive this expression for conditional probability?

    ##P(T=1|W=w)=\frac{P(\{T=1\}\cap\{W=w\})}{P(W=w)}=\frac{\binom {n-2} {w-1} p^{w-1}(1-p)^{(n-2)-(w-1)}}{\binom n w p^w (1-p)^{n-w}}=\frac{(n-2)!}{(w-1)!(n-w-1)!}\frac{w!(n-w)!}{n!}\frac{1}{p(1-p)}=\frac{w(n-w)}{n(n-1)}(p(1-p))^{-1}##. I cannot seem to get the terms with ##p## out of my expression.
  18. C

    MHB Calculating Conditional Probability of Male/Female Customers Buying Books A-D

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  19. H

    Conditional Probability of a continuous joint distribution function

    For 1) I found two ways but I get difference results. The first way is I use P(A|B) = P(A and B)/P(B). I get P(X<1|Y<1)=(∫_0^1▒∫_0^1▒〖3/4 (2-x-y)dydx〗)/(∫_0^1▒∫_0^1▒〖3/4 (2-x-y)dydx〗+∫_1^2▒∫_0^(2-x)▒〖3/4 (2-x-y)dydx〗)=6/7 The 2nd method is I use is f(x│y)=f(x,y)/(f_X (x)...
  20. Vital

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  21. Vital

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  22. N

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  23. N

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  24. D

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  25. L

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  26. R

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  27. M

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  28. Avatrin

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  29. G

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    I don't get $$\frac{P[x<X<x+dx|N=n]}{dx}=f_{X|N}(x|n)$$ Can someone derive why? I would believe that $$f_{X|N}(x|n)=\frac{f(x,N)}{p_n(N)}$$ but I don't get how that would be the same. And I don't get that $$\frac{P[x<X<x+dx|N=n]}{dx}=\frac{P[N=n|x<X<x+dx]}{P[N=n]}\frac{P[x<X<x+dx]}{dx}$$ Can...
  30. Clifford Engle Wirt

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  31. S

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  32. N

    MHB Conditional Probability - Faulty Plumbing

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  33. M

    MHB Conditional probability prove or disprove

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  34. J

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  35. A

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  36. R

    MHB Conditional Probability

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  37. R

    MHB Calculating the Probability of Faulty Plumbing in Hotel Rooms: A Case Study

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  38. W

    I Poisson distribution with conditional probability

    Hi guys, I have a question about computing conditional probabilities of a Poisson distribution. Say we have a Poisson distribution P(X = x) = e^(−λ)(λx)/(x!) where X is some event. My question is how would we compute P(X > x1 | X > x2), or more specifically P(X> x1 ∩ X > x2) with x1 > x2? I...
  39. HaLAA

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  40. J

    MHB What is the probability of a customer only insuring one non-sports car?

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  41. TheSodesa

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  42. R

    I Can Conditional Probability Be Solved Generally with PDFs of Variables?

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  43. Rampart

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  44. SlowThinker

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  45. M

    How Do Different Approaches to Conditional Probability Affect Problem Solving?

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  46. N

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  47. STEMucator

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  48. STEMucator

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  49. Linder88

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