Probability distribution, find constant

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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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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