Python: (Probability) Independent "And" Probability

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Jason
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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}%.")  
\(P_{I}(E_{1} \wedge E_{2}...\wedge\,\,E_{n + 1}) = \)

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

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.
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

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

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

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

\((\dfrac{1}{2}) * (\dfrac{1}{2}) * (\dfrac{1}{2}) * (\dfrac{1}{2}) = \)

\(\dfrac{1}{16} = 0.625 = 62.5\%\)

Given that:

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

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