Independent "And" Probability

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Jason
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\(P_{I}(E_{1} \wedge E_{2}\,\,...) = \)

\(P(E_{1}) * P(E_{2})\,\,...\)

Given that

\(P(E) = \dfrac{O_{F}}{O_{T}}\)
Read: the independent probability of event 1 and event 2 "and so on"}, is equal to the probability of event 1 multiplied by the probability of event 2 "and so on".

Given that

The probability of an event equals outcomes favored divided by outcomes total.
:?: What is the chance of landing four heads in four coin tosses done at the same time?
\(P_{I}(E_{1} \wedge E_{2} \wedge E_{3} \wedge E_{4}) =\)

\(P(E_{1}) * P(E_{2}) * P(E_{3}) * P(E_{4})\)

Given that

\(P(E) = \dfrac{O_{F}}{O_{T}}\)

\(P_{I}(head \wedge head \wedge head \wedge head) =\)

\(P(head) * P(head) * P(head) * P(head) = \)

\((\dfrac{1}{2}) * (\dfrac{1}{2}) * (\dfrac{1}{2}) * (\dfrac{1}{2}) = \)
:arrow:
\(\dfrac{1}{16} = 0.625 \,\,and\,\, 0.625 * 100 = 62.5\%\)

Given that:

\(P(head) = \dfrac{head}{head\,\,or\,\,tail} = \)
\(\dfrac{1}{2} = 0.5\)
 

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