SOLUTION: Let P(x) be the probability function of a discrete random variable x which assumes the values x1, x2, x3, x4, such that 2P(x1) = 3P(x2) = P(x3) = 5P(x4). Find the probability dist

Question 102450: Let P(x) be the probability function of a discrete random variable x which assumes the values x1, x2, x3, x4, such that 2P(x1) = 3P(x2) = P(x3) = 5P(x4). Find the probability distribution of x.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Let P(x) be the probability function of a discrete random variable x which assumes the values x1, x2, x3, x4, such that 2P(x1) = 3P(x2) = P(x3) = 5P(x4). Find the probability distribution of x.
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The least common multiple of 2,3,1,and 5 is 30
So to avoid fractions:
Let P(x3) be 30x
Then:
P(x4)= 6x
P(x2)= 10x
P(x1)= 15x
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Since the probabilities must sum to one you get:
30x + 6x +10x + 15x = 1
61x = 1
x = 1/61
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P(x3)=30/61
P(x4)=6/61
P(x2)=10/61
P(x1)=15/61
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Cheers,
stan H.
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