Question 8.1: A company producing cereals offers a toy in every sixth cere......

The random variable X: “number of packages with a toy” is binomially distributed. In each of n = 20 “trials”, a toy can be found with probability p = \frac{1}{6} .

(a) We thus get

P(X = 4) = \left(\begin{matrix} n \\ k \end{matrix} \right) p^{k}\left(1 − p\right) ^{n-k}= \left(\begin{matrix} 20 \\ 4\end{matrix} \right)\left(\begin{matrix} 1 \\ 6 \end{matrix} \right)^{4}\left(\begin{matrix} 5 \\ 6 \end{matrix} \right)^{16}\approx 0.20.

(b) Similarly, we calculate

P(X = 0) = \left(\begin{matrix} n \\ k \end{matrix} \right) p^{k}\left(1 − p\right) ^{n-k}= \left(\begin{matrix} 20 \\ 0\end{matrix} \right)\left(\begin{matrix} 1 \\ 6 \end{matrix} \right)^{0}\left(\begin{matrix} 5 \\ 6 \end{matrix} \right)^{20}\approx 0.026.

(c) This question relates to a hypergeometric distribution: there are N = 20 packages with M = 3 packages with toys and N − M = 17 packages without a toy. The daughter gets n = 5 packages and we are interested in P(X = 2). Hence, we get

P(X = 2) = \frac{\left(\begin{matrix} M \\ x \end{matrix} \right)\left(\begin{matrix} N-M \\ n-x \end{matrix} \right)}{\left(\begin{matrix} N \\ n \end{matrix} \right)}=\frac{\left(\begin{matrix} 3 \\ 2 \end{matrix} \right)\left(\begin{matrix} 17 \\ 3 \end{matrix} \right)}{\left(\begin{matrix} 20 \\ 5 \end{matrix} \right)} \approx 0.13.