Probability distribution, find constant

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Homework Help Overview

The discussion revolves around a probability distribution problem where the original poster presents a set of probabilities for discrete values of x and seeks to find a constant k in the equation P(x) = k(5-x). Participants are examining the validity of the provided probabilities and their relationship to the proposed equation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question whether the original poster has correctly stated the problem, particularly regarding the sum of the probabilities. Others explore the implications of the equation P(x) = k(5-x) and its compatibility with the given probabilities.

Discussion Status

The discussion is active with participants providing feedback on the original poster's understanding of the problem. There is a recognition that the provided probabilities may not align with the equation, and some participants are seeking clarification on the source of the probabilities listed.

Contextual Notes

There is uncertainty about whether the probabilities were given in the problem statement or derived from the original poster's attempts. The discussion also highlights the necessity for the probabilities to sum to 1 in a valid probability distribution.

Deathfish
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Homework Statement



x = 0, P(x) = 0.4
x = 1, P(x) = 0.1
x = 2, P(x) = 0.1
x = 3, P(x) = 0.1
x = 4, P(x) = 0.3

If P(x)=k(5-x) for x = 0,1,2,3,4, find value of constant k

The Attempt at a Solution



0.4 = k(5-0)
0.1 = k(5-1)
0.1 = k(5-2)
0.1 = k(5-3)
0.3 = k(5-4)

5k+4k+3k+2k+k=1
15k=1
k = 1/15
 
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Deathfish said:
If P(x)=k(5-x) for x = 0,1,2,3,4, find value of constant k

Are you sure you have stated the problem correctly? The sum of the probabilities is already 1.
 
The probability distribution,
x = 0, P(x) = 0.4
x = 1, P(x) = 0.1
x = 2, P(x) = 0.1
x = 3, P(x) = 0.1
x = 4, P(x) = 0.3

does NOT satisfy P(x)= k(5- x). If it did, then you would have to have P(0)= 0.4= k(5)so that k= 0.4/5= 0.08 but then P(1)= k(5-1)= 0.08(4)= 0.32, not 0.1. If P(x)= k(5- x) then 15k= 1 because any probability distribution must sum to 1, not because of "0.4+ 0.1+ 0.1+ 0.1+ 0.3= 1".
 
Deathfish said:

Homework Statement



x = 0, P(x) = 0.4
x = 1, P(x) = 0.1
x = 2, P(x) = 0.1
x = 3, P(x) = 0.1
x = 4, P(x) = 0.3

If P(x)=k(5-x) for x = 0,1,2,3,4, find value of constant k

Deathfish, I have apparently misunderstood your problem. Tell me, are the numbers 0.4,0.1,0.1,0.1, and 0.3 you have listed above supposed to have been somehow given in the statement of the problem or they a result of your attempt at solving the problem? If they are results of your work they shouldn't be stated as part of the problem. If they are the result of your work, they are wrong as Halls has pointed out.
 
0.4,0.1,0.1,0.1, and 0.3 are the values of P(x) in the question... values of x and corresponding P(x) are listed down in a table although i don't know how to post a table here... i have no idea what the question means by "If P(x)=k(5-x) for x = 0,1,2,3,4, find value of constant k" this i am copying down from the question too
 
If P(x)= k(5- x), then it can not be the same P as for the given list.
 

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