\(P(E_{1} \vee E_{2}\,\,...)= \)[Inclusive "Or"] probability means any one of two or more more "or" choices can happen. It's due to the situation of being non-mutually exclusive
\(P(E_{1}) + P(E_{2})\,\,...
\)
\(- P(E_{1} \wedge E_{2}\,\,...)\)
Given that:
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
\(O_{T} \ne 0\)
Read: The probability of event 1 [inclusive "or"] event 2, [inclusive "or"] "and so on" equals the probability of event 1 added to the probability of event 2 "and so on" subtracted by the probability of event 1 and event 2 "and so on".
Given that
The probability of an event is the outcomes favored divided by the outcomes total.
\(P(E_{1} \vee E_{2}) = P(E_{1}) + P(E_{2}) - P(E_{1} \wedge E_{2}) \)What is the chance of unknowingly picking a banana or an orange from a basket containing a banana, orange, apple, pear, kiwi, tangerine, and grapefruit (two picks)? This is assuming any one of both "or" choices are possible.
Given that:
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
\(P(banana \vee orange) = P(banana) + P(orange) - P(banana \wedge orange) \)
\(P(banana \vee orange) = \)
\(\dfrac{1}{7} + \dfrac{1}{7} - P(\dfrac{1}{7} * \dfrac{1}{7}) = \)
\(\dfrac{2}{7} - (\dfrac{1}{49}) = \dfrac{13}{49} = 0.26530612244 = 0.27 \%\)
Given that:
\(P(banana) = \dfrac{banana}{total\,\,fruit} = \dfrac{1}{7}\)
\(P(orange) = \dfrac{orange}{total\,\,fruit} = \dfrac{1}{7}\)
