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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.

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.

Question Image
Grade:11

1 Answers

Manika gupta
askIITians Faculty 43 Points
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)

Add all
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

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