Fill in the blanks in the following table:P(A)P(B)P(A∩B)P(A∪B)(i)\

Fill in the blanks in the following table:P(A)P(B)P(A∩B)P(A∪B)(i)\(\frac{1}{3}\)\(\frac{1}{5}\)\(\frac{1}{15}\) .........(ii)0.35..... 0.250.6(iii)0.50.35.......0.7

Answer»

We need to fill the table:

(i) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Using data from table, we get:

∴ P(A ∪ B) = \(\frac{1}{3}+\frac{1}{5}-\frac{1}{15} = \frac{8}{15}-\frac{1}{15}=\frac{7}{15}\)

(ii) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(B) = P(A ∪ B) + P(A ∩ B) – P(A)

Using data from table, we get:

∴ P(B) = 0.6 + 0.25 – 0.35 = 0.5

(iii) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∩ B) = P(B) + P(A) - P(A ∪ B)

Using data from table, we get:

∴ P(A ∩ B) = 0.5 + 0.35 – 0.7 = 0.15

Filled table is: