Two balanced dice are rolled simultaneously, so each of 36outcomes has the proba

Question

Two balanced dice are rolled simultaneously, so each of 36outcomes has the probability =(1/36). Let X and Y denote the number of spots shwon by first andsecond dice, respectively. Events E and F are defined as follows. E={X+Y=7, or X=Y} F={at least one die shows 6} a) find the joint and marginal probabilites: E Ec Marginal F P(E F)= P(Ec F)= P(F)= Fc P(E Fc)= P(Ec Fc)= P(Fc)= Marginal P(E)= P(Ec)= 1.00
b) Find the conditional propabilities: P(E|F)= P(F|E)= c) Wht is the probability that the union (E or F)occurs? P(E or F) = d) Are events E and F independent? Why? Two balanced dice are rolled simultaneously, so each of 36outcomes has the probability =(1/36). Let X and Y denote the number of spots shwon by first andsecond dice, respectively. Events E and F are defined as follows. E={X+Y=7, or X=Y} F={at least one die shows 6} a) find the joint and marginal probabilites: E Ec Marginal F P(E F)= P(Ec F)= P(F)= Fc P(E Fc)= P(Ec Fc)= P(Fc)= Marginal P(E)= P(Ec)= 1.00 E Ec Marginal F P(E F)= P(Ec F)= P(F)= Fc P(E Fc)= P(Ec Fc)= P(Fc)= Marginal P(E)= P(Ec)= 1.00
b) Find the conditional propabilities: P(E|F)= P(F|E)= c) Wht is the probability that the union (E or F)occurs? P(E or F) = d) Are events E and F independent? Why? E Ec Marginal F P(E F)= P(Ec F)= P(F)= Fc P(E Fc)= P(Ec Fc)= P(Fc)= Marginal P(E)= P(Ec)= 1.00

Explanation / Answer

E Ec Marginal F (1,6) (6,1) (6,6)
(2,6) (3,6) (4,6) (5,6)
(6,2) (6,3) (6,4) (6,5)
3+8=11
Fc (2,5) (3,4) (4,3) (5,2)
(1,1) (2,2) (3,3) (4,4)
(5,5)
(5,4) (1,2) (1,3) (1,4)
(1,5) (2,1) (2,3) (2,4)
(3,1) (3,2) (3,5) (4,1)
(4,2) (4,5) (5,1) (5,3)

9+16=25
Marginal 3+9=12
8+16=24 36
E Ec Marginal F (1,6) (6,1) (6,6)
(2,6) (3,6) (4,6) (5,6)
(6,2) (6,3) (6,4) (6,5)
3+8=11
Fc (2,5) (3,4) (4,3) (5,2)
(1,1) (2,2) (3,3) (4,4)
(5,5)
(5,4) (1,2) (1,3) (1,4)
(1,5) (2,1) (2,3) (2,4)
(3,1) (3,2) (3,5) (4,1)
(4,2) (4,5) (5,1) (5,3)

9+16=25
Marginal 3+9=12
8+16=24 36

b) P(E|F) = 3/11, P(F|E) = 3/12 = 1/4

c) P(E or F) = 1 - P(Ec and Fc) = 1- (16/36)= 20/36 = 5/9

d) P(E) = 12/36 = 1/3
    P(F) = 11/36
    P(E and F if independent) = 1/3 x 11/36 = 11/108 3/36 (1/12)

Thus E and F are not independent.

E Ec Marginal F (1,6) (6,1) (6,6)
(2,6) (3,6) (4,6) (5,6)
(6,2) (6,3) (6,4) (6,5)
3+8=11
Fc (2,5) (3,4) (4,3) (5,2)
(1,1) (2,2) (3,3) (4,4)
(5,5)
(5,4) (1,2) (1,3) (1,4)
(1,5) (2,1) (2,3) (2,4)
(3,1) (3,2) (3,5) (4,1)
(4,2) (4,5) (5,1) (5,3)

9+16=25
Marginal 3+9=12
8+16=24 36

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