Dependent Probability
Posted: Thu Jan 08, 2026 5:54 pm
\(P_{D}(E_{1} \longrightarrow E_{2}\,...) = \)
\(P(E_{1}) * [P(E_{2}\,\, | \,\,E_{1})\, * P(E_{3})\,\,|\,\,E_{2}) ...] \)
Given that:
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
\(O_T \ne 0\)
\(P(E_{1}) * [P(E_{2}\,\, | \,\,E_{1})\, * P(E_{3})\,\,|\,\,E_{2}) ...] \)
\(P_{D}(1\,\,O\,B \longrightarrow 1\,\,G\,B) = \)
\(\,\,and\,\,0.11363636363 * 100 = 11.363636363\% = 11.4\% \)
\(P(E_{1}) * [P(E_{2}\,\, | \,\,E_{1})\, * P(E_{3})\,\,|\,\,E_{2}) ...] \)
Given that:
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
\(O_T \ne 0\)
Read: The dependent probability of event 1 and then event 2 and so on, equals the probability of event 1 multiplied by the probability of event 2, given that event 1 already happened multiplied by the probability of event 3, given that event 2 already happened, and so on.
Given that
The probability of an event equals outcomes favored divided by outcomes total.
Outcomes total cannot equal zero
\(P_{D}(E_{1} \longrightarrow E_{2}\,...) = \)There are 3 green balls, 4 red balls, and 5 orange balls. What is the probability that you picked one orange ball and then you get one green one, assuming nothing is replaced?
\(P(E_{1}) * [P(E_{2}\,\, | \,\,E_{1})\, * P(E_{3})\,\,|\,\,E_{2}) ...] \)
\(P_{D}(1\,\,O\,B \longrightarrow 1\,\,G\,B) = \)
\(\dfrac{5}{12} * [\dfrac{3}{11}] = \dfrac{5}{44} = 0.11363636363\)
\(\,\,and\,\,0.11363636363 * 100 = 11.363636363\% = 11.4\% \)