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8) Let x є { 1, 2, 3, 4, 5, 6} be the outcome of throwing a fair dice. We place two equal weights on the sides 5 and 6 and for our probability law we have the relation P5 or 6)-3P 1 or x-2 or x 3 or x-4) Compute P(z-j) for j є { 1, 2, 3, 4, 5, 6, }

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Answer #1

here as sum of probability is equal to 1

therefore P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=1

P(X=1)+P(X=1)+P(X=1)+P(X=1)+3*P(X=1)+3*P(X=1)=1

P(X=1)=1/10

therefore

P(X=1)=1/10

P(X=2)=1/10

P(X=3)=1/10

P(X=4)=1/10

P(X=5)=3/10

P(X=6)=3/10

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