\(P(E_{1} \veebar E_{2}\,\,...) =\)[Exclusive "Or"] means only one "or" choice can happen. It's connected to being mutually exclusive.
\(P(E_{1}) + P(E_{2})\,\,...\)
Given that
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
\(O_{T} \neq 0\)
Read: The probability of event 1 [exclusive "or"] event 2 [exclusive "or"] "and so on" equals the probability of event 1 added to the probability of event 2 "and so on".
Given that
The probability of an event is the outcomes favored divided by the outcomes total and "outcomes total" cannot equal 0.
\(P(E_{1} \veebar E_{2}) = P(E_{1}) + P(E_{2}) .\)What is the chance of unknowingly picking an apple or an orange from a basket containing an apple, orange and banana? (one pick) This is assuming there is only one correct "or" choice.
Given that:
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
\(P(apple \veebar orange) = P(apple) + P(orange) .\)
\(P(apple \veebar orange) = \)
\(\dfrac{1}{3} + \dfrac{1}{3} = \dfrac{2}{3}\)
\(= 0.667 = 66.7 \%\)
Given that:
\(P(apple) = \dfrac{apple}{total\,\,fruit} = \dfrac{1}{3}\)
\(P(orange) = \dfrac{orange}{total\,\,fruit} = \dfrac{1}{3}\)
