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Let x,y and z be three events such that the probability of occuring exactly one out of x and y is 1-a, out of y and z is 1-2a, out of z and x is 1-3/2a and that of occuring three events simultaneously is a^2. Find the minimum value of the probability that at least one out of x,y,z will occur.

Manika gupta
24 days ago
Dear Student,
You can see your solution here.

P(Exactly one of x or y)
P(x) + P(y)-2P(x^y)
=1-a----------eq (1)
P(y) +P(z)-2P(y^z)
=1-2a---------eq(2)
P(x) +P(z)-2P(z^x)
=1-3/2a-------------eq(3)

2[P(x) + P(y) +P(z)-P(x^y)-P(y^z)-P(z^x)]3-9a/2
P(x) + P(y) +P(z)-P(x^y)-P(y^z)-P(z^x)=3/2-9a/4
P(x^y^z)=a*a
Now atleast of one
P(atleast,X,Y,Z)
P(x) + P(y) +P(z)-P(x^y)-P(y^z)-P(z^x)+P(x^y^z)=3/2 -9/4a +a*a
6-9a+4a*a/4