P(X = x) = 1/19 (1 +|4 - x|)


E(X) = ∑ xP(X = x)

= 1/19 (1 +|4 - 1|)
+ 2(1/19)(1 + (1 +|4 - 2|)
+ 3(1/19)(1 + (1 +|4 - 3|)
+ 4(1/19)(1 + (1 +|4 - 4|)
+ 5(1/19)(1 + (1 +|4 - 5|)
+ 6(1/19)(1 + (1 +|4 - 6|)
+ 7(1/19)(1 + (1 +|4 - 7|)


= 1/19 (1 +|3|)
+ 2(1/19)(1 +|2|)
+ 3(1/19)(1 +|1|)
+ 4(1/19)(1 +|0|)
+ 5(1/19)(1 +|-1|)
+ 6(1/19)(1 +|-2|)
+ 7(1/19)(1 +|-3|)


= 4/19
+ 2(3/19)
+ 3(2/19)
+ 4(1/19)
+ 5(2/19)
+ 6(3/19)
+ 7(4/19)


= 4/19 + 6/19 + 6/19 + 4/19 + 10/19 + 18/19
+ 28/19

= 76/19

= 4

Var(X) = E(X²) - [E(X)]²

= ∑ x²P(X = x) - 4²

= 1²(4/19)
+ 2²(3/19)
+ 3²(2/19)
+ 4²(1/19)
+ 5²(2/19)
+ 6²(3/19)
+ 7²(4/19)
- 16

(From previous result)


= 4/19 + 12/19 + 18/19 + 16/19 + 50/19 + 108/19 + 196/19 - 16

= 404/19 - 304/19

= 100/19

P(X₁+X₂ = 10 | X₁ ≥ 6 )

This means that X₁ can only be either 6 or 7. We basically find all the possible combinations that give a total of 10.


= (P(X₁ = 6 and X₂ = 4) + P(X₁ = 7 and X₂ = 3)) ÷ (P(X₁ = 6) + P(X₁ = 7))

= (P(X = 6) × P(X = 4) + P(X = 7) × P(X = 3)) ÷ (P(X₁ = 6) + P(X₁ = 7))


(Since they are independent so there is no conditional probability within the two observations. Simply multiply their probabilities.)

= (3/19 × 1/19 + 4/19 × 2/19) ÷ (3/19 × 4/19)

= (3/381 + 8/381) ÷ 7/19

= 11/381 ÷ 7/19

= 11/381 × 19/7

= 11/(19×7)

= 11/133

thanks for taking the time to type all of these out!

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