Please show work thanksSolution18 PA 02 002 022 19 PY 00

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Solution

18. P(A) = 0.2 + 0.02 = 0.22

19. P(Y) = 0.02 + 0.08 + 0.02 = 0.12

20. P(A and Y) = 0.02

21. P(A or Y) = P(A) + P(Y) - P(A and Y) = 0.22 + 0.12 - 0.02 = 0.32

22. P(A|Y) = P(A and Y) / P(Y) = 0.02/0.12 = 0.1667

23. P(Y|A) = P(A and Y) / P(A) = 0.02/0.22 = 0.091

24. P(A and Y) = 0.02

If A and Y are independent P(A and Y ) = P(A)*P(Y)

P(A)*P(Y) = 0.22*0.12 = 0.0264

As, they are not equal, A and Y are not independent

25. If A and Y are mutually exclusive, P(A and Y) = 0

But, P(A and Y) is not equal to 0

Therefore, A and Y are not exclusive.

26. Expected Value = 1(0.2) + 2(0.3) + 3(0.05) + 4(0.45) = 2.75

27. Variance = (1-2.9)²*0.1 + (2-2.9)²*0.15 + (3-2.9)²*0.5 + (4-2.9)²*0.25 = 0.79

28. P(Dice = D1|X = 4) = P(Dice = D1 and X = 4) / P(X = 4)

= 0.25*0.25 / (0.25*0.25 + 0.75*0.167)

= 1/3

=0.333