Consider a situation where products are produced sequentially. The events producing defective products are independent and identically distributed, and a defective product is produced with a probability of $\phi$ $(0 \leq \phi \leq 1)$. Suppose that the probability of producing a defective product follows the Beta distribution $$\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(x) = \frac{1}{\mathrm{B}(\mathrm{a},\mathrm{b})} x^{\mathrm{a}-1}(1-x)^{\mathrm{b}-1} \quad (0 \leq x \leq 1),$$ for real numbers $\mathrm{a}(>1)$ and $\mathrm{b}(>1)$. Note that the Beta function $\mathrm{B}(\mathrm{a},\mathrm{b})$ is defined as $$\mathrm{B}(\mathrm{a},\mathrm{b}) = \int_0^1 t^{\mathrm{a}-1}(1-t)^{\mathrm{b}-1} \mathrm{d}t$$ In Question II.4, where the posterior probability is the Beta distribution $\operatorname{Beta}_{\mathrm{a}^{\prime},\mathrm{b}^{\prime}}(\phi)$ with $a=2,\ b=50$, obtain $\phi$ that gives the maximum likelihood estimate (that maximizes the posterior probability).
Consider a situation where products are produced sequentially. The events producing defective products are independent and identically distributed, and a defective product is produced with a probability of $\phi$ $(0 \leq \phi \leq 1)$. Suppose that the probability of producing a defective product follows the Beta distribution
$$\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(x) = \frac{1}{\mathrm{B}(\mathrm{a},\mathrm{b})} x^{\mathrm{a}-1}(1-x)^{\mathrm{b}-1} \quad (0 \leq x \leq 1),$$
for real numbers $\mathrm{a}(>1)$ and $\mathrm{b}(>1)$. Note that the Beta function $\mathrm{B}(\mathrm{a},\mathrm{b})$ is defined as
$$\mathrm{B}(\mathrm{a},\mathrm{b}) = \int_0^1 t^{\mathrm{a}-1}(1-t)^{\mathrm{b}-1} \mathrm{d}t$$
In Question II.4, where the posterior probability is the Beta distribution $\operatorname{Beta}_{\mathrm{a}^{\prime},\mathrm{b}^{\prime}}(\phi)$ with $a=2,\ b=50$, obtain $\phi$ that gives the maximum likelihood estimate (that maximizes the posterior probability).