Aufgabe:
… Given that there were 12 heads in 20 independent coin tosses, calculate
a) the chance that the first toss landed heads;
b) the chance that the first two tosses landed heads;
c) the chance that at least two of the first five tosses landed heads.
Problem/Ansatz:
… Benötige bei allen Aufgaben Hilfe, aber das hier wäre ein Ansatz, der vermutlich nicht ganz richtig ist für c)
\( P(B)=P(B=2)+P(B=3)+P(B=4)+P(B=5) \)
\( =\left(\begin{array}{l}5 \\ 2\end{array}\right)\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{3}+\left(\begin{array}{l}5 \\ 3\end{array}\right)\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{2}+\left(\begin{array}{c}5 \\ 4\end{array}\right)\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{1}+\left(\begin{array}{l}5 \\ 5\end{array}\right)\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{0} \)
\( =\left(\begin{array}{l}5 \\ 2\end{array}\right)\left(\frac{1}{2}\right)^{5}+\left(\begin{array}{l}5 \\ 3\end{array}\right)\left(\frac{1}{2}\right)^{5}+\left(\begin{array}{c}5 \\ 4\end{array}\right)\left(\frac{1}{2}\right)^{5}+\left(\begin{array}{c}5 \\ 5\end{array}\right)\left(\frac{1}{2}\right)^{5} \)
\( P(B \mid A)=\sum \limits_{k=2}^{n=5} \frac{P(A \mid B=k) P(B=k)}{P(A)}=\frac{\left(\begin{array}{l}18 \\ 10\end{array}\right)\left(\begin{array}{l}5 \\ 2\end{array}\right)+\left(\begin{array}{c}17 \\ 9\end{array}\right)\left(\begin{array}{l}5 \\ 3\end{array}\right)+\left(\begin{array}{c}16 \\ 8\end{array}\right)\left(\begin{array}{l}5 \\ 4\end{array}\right)+\left(\begin{array}{c}15 \\ 7\end{array}\right)\left(\begin{array}{l}5 \\ 5\end{array}\right)}{\left(\begin{array}{c}20 \\ 1\end{array}\right)} \)
