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 X 1 2 3 4 5 6
 P(X=x) k 2k 3k 4k 5k 6k
The probability distribution of discrete r.v. $$X$$ is 
Then $$P(X\le 4)$$ = 


Solution

Sum of the probabilities is equal to 1

$$\therefore k+2k+3k+4k+5k+6k=1$$

$$21k=1$$

$$k=\dfrac{1}{21}$$

$$P(X\leq 4)$$

$$=P(1)+P(2)+P(3)+P(4)$$

$$=k+2k+3k+4k$$

$$=\dfrac{1}{21}+\dfrac{2}{21}+\dfrac{3}{21}+\dfrac{4}{21}$$

$$=\dfrac{1+2+3+4}{21}$$

$$=\dfrac{10}{21}$$

$$\therefore P(X\leq 4)=0.476$$

Applied Mathematics

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