In a bombing attack, there is $50\%$ chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that the probability of the target being destroyed is at least $0.99$, is ...

The coefficients $a , b , c$ in the quadratic equation $a x ^ { 2 } + b x + c = 0$ are from the set $\{ 1,2,3,4,5,6 \}$. If the probability of this equation having one real root bigger than the other is $p$, then 216 p equals :
(1) 57
(2) 76
(3) 38
(4) 19

A research team uses a certain rapid test reagent to understand the proportion of organisms in a protected area whose body toxin accumulation exceeds the standard due to environmental pollution. The test result of this reagent shows only two colors: red and yellow. Based on past experience, it is known that: if body toxin accumulation exceeds the standard, after testing with this reagent, $75\%$ shows red; if body toxin accumulation does not exceed the standard, after testing with this reagent, $95\%$ shows yellow. It is known that for a certain type of organism in this protected area, $7.8\%$ of the test results show red. Assuming the actual proportion of this type of organism with body toxin accumulation exceeding the standard is $p\%$ , select the correct option.
(1) $1 \leq p < 3$
(2) $3 \leq p < 5$
(3) $5 \leq p < 7$
(4) $7 \leq p < 9$
(5) $9 \leq p < 11$

An electronics company has several hundred employees with two types of meal arrangements: bringing meals from home or eating out. Long-term surveys have found that: if an employee brings meals from home on a given day, then $10\%$ will switch to eating out the next day; if an employee eats out on a given day, then $20\%$ will switch to bringing meals from home the next day. Let $x _ { 0 }$、$y _ { 0 }$ respectively represent the proportion of employees bringing meals from home and eating out today relative to the total number of employees, where $x _ { 0 }$、$y _ { 0 }$ are both positive, and $x _ { n }$、$y _ { n }$ respectively represent the proportion of employees bringing meals from home and eating out after $n$ days relative to the total number of employees. Given that the number of employees in the company remains unchanged, select the correct options.
(1) $y _ { 1 } = 0.9 y _ { 0 } + 0.2 x _ { 0 }$
(2) $\left[ \begin{array} { l } x _ { n + 1 } \\ y _ { n + 1 } \end{array} \right] = \left[ \begin{array} { l l } 0.9 & 0.2 \\ 0.1 & 0.8 \end{array} \right] \left[ \begin{array} { l } x _ { n } \\ y _ { n } \end{array} \right]$
(3) If $\frac { x _ { 0 } } { y _ { 0 } } = \frac { 2 } { 1 }$ , then $\frac { x _ { n } } { y _ { n } } = \frac { 2 } { 1 }$ holds for any positive integer $n$
(4) If $y _ { 0 } > x _ { 0 }$ , then $y _ { 1 } > x _ { 1 }$
(5) If $x _ { 0 } > y _ { 0 }$ , then $x _ { 0 } > x _ { 1 }$

It is known that 30\% of the population in a certain region is infected with a certain infectious disease. For a rapid screening test of the disease, there are two results: positive or negative. The test has an 80\% probability of identifying an infected person as positive and a 60\% probability of identifying an uninfected person as negative. To reduce the situation where the test incorrectly identifies an infected person as negative, experts recommend three consecutive tests. If $P$ is the probability that an infected person is among those who test negative in a single test, and $P'$ is the probability that an infected person is among those who test negative in all three consecutive tests, what is $\frac { P } { P' }$ closest to?
(1) 7
(2) 8
(3) 9
(4) 10
(5) 11

A store sells a popular action figure through a lottery. Each lottery draw is independent with a probability of winning of $\frac{2}{5}$. Participants can participate in the lottery using one of the following two methods.
Method 1: Pay 225 yuan to get two lottery chances. Stop drawing as soon as you win and receive one action figure. If you fail to win in both draws, you must pay an additional 75 yuan to receive one action figure.
Method 2: Unlimited number of lottery draws, paying 100 yuan per draw.
If using Method 1 to participate in the lottery, what is the probability of paying a total of 300 yuan to obtain one action figure? (Single choice question, 2 points)
(1) $\left(\frac{2}{5}\right)^{2}$
(2) $\left(\frac{2}{5}\right)^{3}$
(3) $\left(\frac{3}{5}\right)^{2}$
(4) $\left(\frac{3}{5}\right)^{3}$
(5) $\left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right)^{2}$

A store sells a popular action figure through a lottery. Each lottery draw is independent with a probability of winning of $\frac{2}{5}$. Participants can participate in the lottery using one of the following two methods.
Method 1: Pay 225 yuan to get two lottery chances. Stop drawing as soon as you win and receive one action figure. If you fail to win in both draws, you must pay an additional 75 yuan to receive one action figure.
Method 2: Unlimited number of lottery draws, paying 100 yuan per draw.
If using Method 2 to participate in the lottery until winning one action figure, express the expected value of the number of lottery draws needed using the definition of expected value and the $\sum$ notation, and find its value. (Non-multiple choice question, 4 points)

A store sells a popular action figure through a lottery. Each lottery draw is independent with a probability of winning of $\frac{2}{5}$. Participants can participate in the lottery using one of the following two methods.
Method 1: Pay 225 yuan to get two lottery chances. Stop drawing as soon as you win and receive one action figure. If you fail to win in both draws, you must pay an additional 75 yuan to receive one action figure.
Method 2: Unlimited number of lottery draws, paying 100 yuan per draw.
Assuming there is no limit on spending until obtaining one action figure, find the expected value of the amount paid to obtain one action figure using each of the two lottery methods, and explain the relationship between these two expected values. (Non-multiple choice question, 6 points)

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 $\phi$ $(0 \leq \phi \leq 1)$. We consider the changes of the probability distributions before and after observing production results. In the following questions, $N (\geq 1)$ denotes the number of products observed.
By defining $v_i = 1$ if the $i$-th product is a defective product, and $v_i = 0$ if it is not defective, we get a series $\boldsymbol{v} = (v_1, \cdots, v_N)$, where the values can be 0 or 1. Let $N_d(\boldsymbol{v})$ be the number of observations with value of 1 in $\boldsymbol{v}$, obtain the occurrence probability of $\boldsymbol{v}$.

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 the Bayesian estimation, the parameter $\theta$ (in this case, $\phi$) that determines the probability is treated as the random variable and we assume that its distribution is described by $\pi(\theta)$. We calculate $\pi(\theta \mid A)$ by
$$\pi(\theta \mid A) = \frac{\pi(\theta) P(A \mid \theta)}{P(A)}$$
where $\pi(\theta \mid A)$ is the posterior probability, $P(A \mid \theta)$ is the conditional occurrence probability that the event $A$ is observed under $\theta$, and $\pi(\theta)$ is the prior probability.
We assume that $\phi$, the probability of producing a defective product, follows the prior probability $\operatorname{Beta}_{\mathrm{a},\mathrm{b}}(\phi)$. Let $Q(\boldsymbol{v} \mid \phi)$ be the conditional occurrence probability of $\boldsymbol{v}$ under $\phi$ and $Q_{\mathrm{a},\mathrm{b}}(\boldsymbol{v})$ be the occurrence probability of $\boldsymbol{v}$. Obtain the posterior probability after $\boldsymbol{v}$ occurs.

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
$$\mathrm{B}(\mathrm{a},\mathrm{b}) = \int_0^1 t^{\mathrm{a}-1}(1-t)^{\mathrm{b}-1} \mathrm{d}t$$
By defining $v_i = 1$ if the $i$-th product is a defective product, and $v_i = 0$ if it is not defective, we get a series $\boldsymbol{v} = (v_1, \cdots, v_N)$, where the values can be 0 or 1. Let $N_d(\boldsymbol{v})$ be the number of observations with value of 1 in $\boldsymbol{v}$.
Suppose that $Q(\boldsymbol{v} \mid \phi)$ in Question II.2 is the occurrence probability obtained in Question II.1 and let $a = 2,\ b = 50$, obtain $Q_{2,50}(\boldsymbol{v})$.

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.3, with $a=2,\ b=50$, show that the posterior probability becomes the Beta distribution $\operatorname{Beta}_{\mathrm{a}^{\prime},\mathrm{b}^{\prime}}(\phi)$, and obtain $\mathrm{a}^{\prime}$ and $\mathrm{b}^{\prime}$.

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).