Q38 of 43 Page 15

(ii) P(not E and not F).


(i) P(E or F) = P(E ∪ F)

We have, P(E ∪ F) = P(E) + P(F) – P(E and F)

                          =

                          =

                          = =

(ii) P(not E and not F) = P(E’ ∩ F’)

Also E’ ∩ F’ = (E ∩ F)’

(by De Morgan’s Law)

Hence, P(E’ ∩ F’) = P(E ∪ F)’

                        = 1 – P(E ∪ F)’

                        = 1 - = =

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