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Questions tagged [conditional-probability]

For questions on conditional probability.

Conditional probability is the probability that an event occurs given that another event has already happened. The probability of an event $A$ given another event $B$ is written as $P(A|B)$, and is related to the marginal and joint probabilities via $$ P(A|B)P(B)=P(A\cap B)$$

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what is the probability that at least one does not work?

Suppose that you just received a shipment of six televisions and two are defective. If two televisions are randomly selected, compute the probability that both televisions work. here is my work and steps to solve it I believe you should set up a…
23Jerry8Kob
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conditional probability of a statistic given a efficient statistic

Suppose we have random samples $X = (X_1, ...,X_n), $ $X_i \sim iid f(x|\theta)$. $T = T(X)$ is a sufficient statistic for $\theta$. By the definition of sufficiency, we know that $f(x|T=t,\theta) = f(x|T=t)$. Does the following statement hold?…
mxdxzxyjzx
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Conditional Probability with cards

I am struggling to grasp conditional probability as it is something I never fully understood in school. Could anyone explain to me how to answer the following questions? I know how to do a few of them but still confused by the others. Two cards are…
user731762
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P$(A \vert B) = \frac{P(AB)}{P(B)}$ if $P(B) \neq 0$?

I am familiar with the definition $P(A \vert B) = \begin{cases} \dfrac{P(B \vert A) P(A)}{P(B)}, & P(B) \neq 0 \\ \text{undefined}, & P(B) = 0 \end{cases}$ However, I just saw another definition for $P(A \vert B)$: $P(A \vert B) = \begin{cases}…
Dom Fomello
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How do I write this conditional probability proof?

I've been looking at this proof for a while, and I think I was able to answer the first part of the question. It asks: (i) For any events $X$ and $Y$ with $\textbf{P}(Y) > 0$, show that $\textbf{P}(X|Y) + P(X^{c}|Y) = 1$. My proof is as…
TheFiveHundredYears
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Suppose that $A$ and $B$ are disjoint, so $P (A \cap B) = 0$. It follows that $P(A \mid B) = P(B \mid A) = 0$?

My textbook, Statistical Inference by Casella and Berger, says the following: Suppose that $A$ and $B$ are disjoint, so $P (A \cap B) = 0$. It follows that $P(A \mid B) = P(B \mid A) = 0$. Intuitively, I don't see how this makes sense. We know…
The Pointer
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Probability that a patient is infected if test result says so with 0.8 probability

Patients of a clinic are tested for a particular desease. For each patient, the result of the test – ‘infected’/’not infected’ – is correct with the probability 0.8. Suppose that 20% of the patients are infected. What is the probability that a given…
Ridwan Salahuddeen
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Tricky tree diagram probability problem.

$A$ always goes to work by a bus or a taxi. If he goes to work by a bus one day, the probability he goes to work by a taxi the next day is $0.4$. If he goes to work by a taxi one day, the probability he goes to work by a bus the next day is…
Anthony
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Conditional Probability: "Transformation" laws?

I'm struggling to find the right terms for what I mean (not a native speaker), maybe the word "transformation" is wrong in this context? I'm reading a paper which shows the following transformation / formula: I'm having difficulties to understand…
Robert
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Conditional probability question 4 valves

I need help with a question that has confused me. picture of the system : A water flow system has 4 valves For the water to flow A and D must be open and also at least one of the valves B and C. The probability that a particular valve will be open…
Roee
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Question about conditional probability.

This is a probability textbook, and this is about conditional probability. I don't understand this example 1.3-4, specifically that I don't understand how they calculate P(A|B). What's the logic behind?
Beacon
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If it is known that 4 cards have different suits, what is the probability that the first card is a face card? Without replacement.

I'm assuming this is a conditional probability question. Because the four cards are of different suits, we have 52 choices for the first card. Then whatever the first card is, throw away the remaining cards of that suit. In a similar fashion, there…
Chesso
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Is $P(A 丨 B) = P(C 丨 B) + P(D 丨 B) + P(E 丨 B)$ If C, D and E are the subsets of A, and $P(A) =P(C) + P(D)+P(E)$

I was trying to solve this problem. There are three roads that connected town A to town B. road 1 and road 2 have 2 sections, and road 3 has 1 section. The probability that a section will be blocked will bee equal to p, and 0 The road would be…
Henry Cai
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Conditional probability with word play

I have been working on this question for a while and I haven't obtained any reasonable results: In a city, 70% of the inhabitants are non-smokers. Specialists estimate that there is a 45% chance that smokers will suffer from lung cancer at some…
arin
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