Python: (Probability) Independent "And" Probability
Posted: Mon Sep 29, 2025 6:55 pm
Code: Select all
print(f " 8-) What is the chance of landing four heads in four coin tosses done at the same time?")
P_E_1 = P_E_2 = P_E_3 = P_E_4 = 0.5
I_A_P_of_four_head_tosses_on_4_C = P_E_1 * P_E_2 * P_E_3 * P_E_4
percent_I_A_P_of_four_head_tosses_on_4_C = I_A_P_four_head_tosses_on_4_C * 100
print(f "The probability of four head tosses at the same time is {I_A_P_of_four_head_tosses_on_4_C} which is {percent_I_A_P_of_four_head_tosses_on_4_C}%.")
\(P(E_{1}) * P(E_{2})...* \,\, E_{n + 1}\)
Given that
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
Read: the independent probability of event 1 and event 2 and event E_{n + 1}, is equal to the probability of event 1 multiplied by the probability of event 2 multiplied by the probability of event E_{n + 1}.
Given that
The probability of an event equals outcomes favored divided by outcomes total.
\(P_{I}(E_{1} \wedge E_{2} \wedge E_{3} \wedge E_{4}) =\)What is the chance of landing four heads in four coin tosses done at the same time?
\(P(E_{1}) * P(E_{2}) * P(E_{3}) * P(E_{4})\)
Given that
\(P(E) = \dfrac{O_{F}}{O_{T}}\)
\(P_{I}(head \wedge head \wedge head \wedge head) =\)
\(P(head) * P(head) * P(head) * P(head) = \)
\((\dfrac{1}{2}) * (\dfrac{1}{2}) * (\dfrac{1}{2}) * (\dfrac{1}{2}) = \)
\(\dfrac{1}{16} = 0.625 = 62.5\%\)
Given that:
\(P(head) = \dfrac{head}{head\,\,or\,\,tail} = \)
\(\dfrac{1}{2} = 0.5\)