1. **Stating the problem:** We have three types of pastries with given total counts and counts made from quality ingredients. We want to find probabilities related to quality and type. 2. **Given data:** - Cinnamon rolls: total 50, quality 30, non-quality 20 - Chocolate cakes: total 60, quality 40, non-quality 20 - Blueberry buns: total 40, quality 30, non-quality 10 3. **Total pastries:** $50 + 60 + 40 = 150$ 4. **Total quality pastries:** $30 + 40 + 30 = 100$ 5. **Total non-quality pastries:** $20 + 20 + 10 = 50$ --- **a) Probability pastry is non-quality given it is a blueberry bun:** - Total blueberry buns: 40 - Non-quality blueberry buns: 10 $$P(\text{non-quality} | \text{blueberry}) = \frac{10}{40} = 0.25$$ --- **b) Probability pastry is quality (no condition):** $$P(\text{quality}) = \frac{100}{150} = \frac{2}{3} \approx 0.6667$$ --- **c) Probability pastry is chocolate given it is quality:** - Quality chocolate cakes: 40 - Total quality pastries: 100 $$P(\text{chocolate} | \text{quality}) = \frac{40}{100} = 0.4$$ --- **Příklad 2:** Given density function: $$f(x) = \begin{cases} c - \frac{x}{2}, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases}$$ **a) Find constant $c$:** Since $f(x)$ is a pdf, total area under curve must be 1: $$\int_0^2 \left(c - \frac{x}{2}\right) dx = 1$$ Calculate integral: $$\int_0^2 c dx - \int_0^2 \frac{x}{2} dx = 2c - \frac{1}{2} \cdot \frac{2^2}{2} = 2c - 1$$ Set equal to 1: $$2c - 1 = 1 \implies 2c = 2 \implies c = 1$$ **b) Distribution function $F(x)$:** For $x < 0$, $F(x) = 0$. For $0 \leq x \leq 2$: $$F(x) = \int_0^x (1 - \frac{t}{2}) dt = \left[t - \frac{t^2}{4}\right]_0^x = x - \frac{x^2}{4}$$ For $x > 2$, $F(x) = 1$. **c) Expected value $E[X]$ and variance $D[X]$:** $$E[X] = \int_0^2 x (1 - \frac{x}{2}) dx = \int_0^2 (x - \frac{x^2}{2}) dx = \left[\frac{x^2}{2} - \frac{x^3}{6}\right]_0^2 = (2 - \frac{8}{6}) = 2 - \frac{4}{3} = \frac{2}{3}$$ $$E[X^2] = \int_0^2 x^2 (1 - \frac{x}{2}) dx = \int_0^2 (x^2 - \frac{x^3}{2}) dx = \left[\frac{x^3}{3} - \frac{x^4}{8}\right]_0^2 = \left(\frac{8}{3} - \frac{16}{8}\right) = \frac{8}{3} - 2 = \frac{2}{3}$$ Variance: $$D[X] = E[X^2] - (E[X])^2 = \frac{2}{3} - \left(\frac{2}{3}\right)^2 = \frac{2}{3} - \frac{4}{9} = \frac{6}{9} - \frac{4}{9} = \frac{2}{9}$$ **d) Probabilities:** - $P(X=0.5) = 0$ for continuous variables. - $P(X > 1) = 1 - F(1) = 1 - \left(1 - \frac{1^2}{4}\right) = 1 - \left(1 - \frac{1}{4}\right) = \frac{1}{4}$ --- **Příklad 3:** Balls: 5 white, 10 black, total 15. Sampling with replacement 4 times. Probability white ball each draw: $p = \frac{5}{15} = \frac{1}{3}$ Number of white balls in 4 draws is binomial $B(n=4, p=\frac{1}{3})$. **a) Exactly one white ball:** $$P(X=1) = \binom{4}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81} \approx 0.3951$$ **b) At least three white balls:** $$P(X \geq 3) = P(X=3) + P(X=4)$$ Calculate: $$P(X=3) = \binom{4}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 = 4 \cdot \frac{1}{27} \cdot \frac{2}{3} = \frac{8}{81}$$ $$P(X=4) = \binom{4}{4} \left(\frac{1}{3}\right)^4 = 1 \cdot \frac{1}{81} = \frac{1}{81}$$ Sum: $$\frac{8}{81} + \frac{1}{81} = \frac{9}{81} = \frac{1}{9} \approx 0.1111$$ **c) At most two white balls:** $$P(X \leq 2) = 1 - P(X \geq 3) = 1 - \frac{1}{9} = \frac{8}{9} \approx 0.8889$$