Help with conditional probability

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Homework Help Overview

The discussion revolves around conditional probability involving integers selected from a range of 10 to 50. Participants are exploring specific probabilities related to even numbers, numbers greater than 40, and prime numbers within a defined subset.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to calculate conditional probabilities using various interpretations of the problem. Some express confusion about the correct formulas and the application of probability rules. Others provide examples to clarify the concept of conditional probability.

Discussion Status

The discussion includes attempts to clarify misunderstandings regarding the calculation of conditional probabilities. Some participants have offered corrections to the original poster's approach, suggesting a reevaluation of how probabilities are defined and calculated in the context of the problem.

Contextual Notes

There is a noted uncertainty regarding the interpretation of the range for prime numbers, particularly in the context of the phrase "between 20 and 40." Additionally, participants express feelings of frustration and confusion about the concepts involved.

imjello
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A number is selected randomly from a container containing all the integers from 10 to 50 find

a) p(even|greater than 40)
b) p(greater than 40| even)
c) p(prime| between 20 and 40)

please provide an explanation, thanks a lot =D
 
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Idk if people haven't answered because i didn't show work or not but there isn't much work to show in the first place here's what i tried anyway.

a. probability its even .5 (5/10)/ probability its greater than 40(1/5) .2 = 2.5 which is over one so obviously wrong

b. probability its greater than 40(1/5) .2 / .5= .4

c. (26/50).52/ (20/50) .4 = 1.3 once again over one

i feel stupid for not understanding how to do these i understand when its in a chart but can't get these question
 
I'll give an example: say we wanted to find P(odd | less than or equal to 20). This probability is just the proportion of numbers between 10 and 20 (inclusive) that are odd. These numbers are 11,13,15,17,19; there are 5 of them out of 11 choices, so the the probability sought is 5/11.
 
imjello said:
a. probability its even .5 (5/10)/ probability its greater than 40(1/5) .2 = 2.5 which is over one so obviously wrong
Looks like you're thinking that
P(even|greater than 40) = [itex]\frac{P(\text{even})}{P(\text{> 40})}[/itex]
... and that is wrong. It should be
P(even|greater than 40) = [itex]\frac{P(\text{even AND > 40})}{P(\text{> 40})}[/itex]
Try it again.
imjello said:
b. probability its greater than 40(1/5) .2 / .5= .4
Nope. Like in part a, the setup would be
P(greater than 40| even) = [itex]\frac{P(\text{> 40 AND even})}{P(\text{even})}[/itex]
 
imjello said:
A number is selected randomly from a container containing all the integers from 10 to 50 find

a) p(even|greater than 40)
There are 10 numbers in the container that are larger than 40. How many of them are even?

b) p(greater than 40| even)
There are 21 numbers in the container that are even. How many of them are greater than 40?

c) p(prime| between 20 and 40)
There are 19 numbers in the container between 20 and 40. How many of them are prime?
(I am assuming that "between 20 and 40" means 21 to 39.

please provide an explanation, thanks a lot =D
 

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