Answer 1)
chances of solving the problem for A = p(A) = 1/2
chances of solving the problem for B = p(B) = 3/4
chances of solving the problem for C = p(C) = 1/4
i) Probability that the problem will be solved = P(AUBUC)
= P(A) + P(B) + P(C) - P(AB) - P(BC) - P(CA) + P(ABC)
= 1/2 + 3/4 + 1/4 - ((1/2)*(3/4)) - ((3/4)*(1/4)) - ((1/4)*(1/2)) + ((1/2)*(3/4)*(1/4))
= 6/4 - 3/8 - 3/16 - 1/8 + 3/32
= (48 - 12 - 6 - 4 + 3) / 32
= 29/32 = 0.90625
ii)
p(not A) = 1- p(A) =1 -1/2 = 1/2
p(not B) = 1- p(B) =1 -3/4 = 1/4
p(not C) = 1- p(C) =1 -1/4 = 3/4
Probability exactly one of them will solve = P(A)*P(not B)*P(not
C) + P(not A)*P(B)*P(not C)+ P(not A)*P(not B)*P(C)
= (1/2)*(1/4)*(3/4) + (1/2)*(3/4)*(3/4) + (1/2)*(1/4)*(1/4)
= (3/32) + (9/32) + (1/32)
= 13/32 = 0.40625
Answer 2)
P(A) = 0.4
P(B) = p
P(AUB) = 0.7
a)
if A and B are mutually exclusive then
P(AUB) = P(A) + P(B)
0.7 = 0.4 + p
p = 0.7 -0.4 = 0.3
b)
if A and B are independent
P(A ∩ B) = P(A)*P(B)
and P(AUB) = P(A) + P(B) - P(A ∩ B)
0.7 = 0.4 + p - 0.4*p
p = (0.7 - 0.4) / 0.6
p = 0.5
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