Python: (Probability) Exclusive "Or" Probability
Posted: Thu Oct 02, 2025 8:25 pm
[Exclusive "Or"] means one event could happen or another, and so on, but all events cannot happen. It's connected to being mutually exclusive.
Code: Select all
print(f "What is the chance of unknowingly picking an apple or an orange from a basket with an apple, orange, and banana (one fruit pick)?")
float(apple_outcome) = float(orange_,outcome) = float(banana_outcome) = 1/3
apple_or_orange_outcome = apple_outcome + orange_outcome
percent_apple_or_orange_outcome = apple_or_orange_outcome * 100
print(f "The chance of picking an orange or apple is {apple_or_orange_outcome} which is {percent_apple_or_orange_outcome}%")
\(P(E_{1}) + P(E_{2})...+\,\,E_{n + 1}\)
Given that
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
Read: The probability of event 1 [exclusive "or"] event 2 [exclusive "or"] event E_{n+1} equals the probability of event 1 added to the probability of event 2 added to the probability of event E_{n +1}
Given that
The probability of an event is the outcomes favored divided by the outcomes total.
\(P(E_{1} \veebar E_{2}) = P(E_{1}) + P(E_{2}) .\)What is the chance of unknowingly picking an apple or an orange from a basket containing an apple, orange and banana. (one pick)
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:
\(P(apple) = \dfrac{apple}{total\,\,fruit} = \dfrac{1}{3}\)
\(P(orange) = \dfrac{orange}{total\,\,fruit} = \dfrac{1}{3}\)