Question

If P(A)=1/3, P(B)=2/3 and P(A intersection B)=1/6 then find P(A' intersection B')​

Answer 1

Answer:

P(A' INTERSECTION B')= 1/6

Step-by-step explanation:

P(A)= 1/3

P(B)= 2/3

P(A INTERSECTION B)=1/6

P(A' INTERSECTION B')= P(A UNION B)' { DE-Morgan's Law)

P(A UNION B)= P(A) + P(B) - P(A INTERSECTION B)

P(A UNION B)= 1/3 +2/3 -1/6

P(A UNION B)= (2+4-1)/6 = 5/6

P(A UNION B)'= 1-5/6 = 1/6

P(A' INTERSECTION B') = 1/6

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