凯（Kai）和里奥（Leo）两兄弟是定期训练的双项全能（跑步和骑行）运动员。在每一天中，凯骑行的概率为 $\frac{3}{4}$，跑步的概率为 $\frac{3}{5}$，两者都进行的概率为 $p$。 平均而言，里奥在一周的 7 天中有 5 天骑行。在凯骑行的前提下，里奥骑行的概率为 0.9。

1. Express the probability of exactly one event occurring in terms of the given probabilities.

Let $C$ be the event that Kai cycles and $R$ be the event that Kai runs.
$P(\text{either runs or cycles but not both}) = P(C \cap R') + P(C' \cap R)$
$= (P(C) - P(C \cap R)) + (P(R) - P(C \cap R))$

2. Substitute the given values and simplify the expression.

$= \left(\frac{3}{4} - p\right) + \left(\frac{3}{5} - p\right)$
$= \frac{15}{20} + \frac{12}{20} - 2p$
$= \frac{27}{20} - 2p$

1. Determine the lower bound of $p$ using the fact that the maximum probability of the union of two events is 1.

$\mathrm{P}(C \cup R) \le 1$
$\mathrm{P}(C) + \mathrm{P}(R) - \mathrm{P}(C \cap R) \le 1$
$\frac{3}{4} + \frac{3}{5} - p \le 1$
$\frac{27}{20} - p \le 1$
$p \ge \frac{7}{20}$

2. Determine the upper bound of $p$ by considering that the intersection of two events cannot exceed the probability of either individual event, and state the combined range.

$\mathrm{P}(C \cap R) \le \mathrm{P}(C) \quad \text{and} \quad \mathrm{P}(C \cap R) \le \mathrm{P}(R)$
$p \le \frac{3}{4} \quad \text{and} \quad p \le \frac{3}{5}$
$\therefore p \le \frac{3}{5}$
Combining the two inequalities gives the range of possible values for $p$:
$\frac{7}{20} \le p \le \frac{3}{5}$

1. Set up an equation for the total probability that Leo cycles.

Let $K_C$ be the event that Kai cycles and $L_C$ be the event that Leo cycles.
Using the law of total probability,
$\mathrm{P}(L_C) = \mathrm{P}(L_C \cap K_C) + \mathrm{P}(L_C \cap K'_C)$

2. Calculate the probability that Leo cycles and Kai does not.

We are given $\mathrm{P}(L_C) = \frac{5}{7}$, $\mathrm{P}(K_C) = \frac{3}{4}$, and $\mathrm{P}(L_C \mid K_C) = 0.9$.
$\mathrm{P}(L_C \cap K_C) = \mathrm{P}(L_C \mid K_C) \mathrm{P}(K_C) = 0.9 \times \frac{3}{4} = \frac{27}{40}$
Therefore,
$\frac{5}{7} = \frac{27}{40} + \mathrm{P}(L_C \cap K'_C)$
$\mathrm{P}(L_C \cap K'_C) = \frac{5}{7} - \frac{27}{40} = \frac{200 - 189}{280} = \frac{11}{280}$

3. Calculate the required conditional probability.

$\mathrm{P}(K'_C) = 1 - \mathrm{P}(K_C) = 1 - \frac{3}{4} = \frac{1}{4}$
The probability that Leo cycles when Kai does not is:
$\mathrm{P}(L_C \mid K'_C) = \frac{\mathrm{P}(L_C \cap K'_C)}{\mathrm{P}(K'_C)} = \frac{\frac{11}{280}}{\frac{1}{4}} = \frac{11}{70}$

1. State the independence assumption in context.

Whether Leo cycles on any particular day is independent of whether he cycles on any other day.

2. State the constant probability assumption in context.

The probability that Leo cycles is the same for each day.

1. Define the random variable and its binomial distribution.

Let $X$ be the number of days Leo cycles in $5$ weeks.
Since there are $7$ days in a week, the total number of days is $n = 5 \times 7 = 35$.
Based on his average, the probability that Leo cycles on any given day is $p = \frac{5}{7}$.
Therefore, we model this as a binomial distribution:
$X \sim \mathrm{B}\left(35, \frac{5}{7}\right)$

2. Formulate the required probability interval and compute it using a graphing calculator.

We need to find the probability that Leo cycles on at least $20$ days but fewer than $30$ days. Since $X$ is a discrete random variable, "fewer than $30$" means a maximum of $29$.
$\mathrm{P}(20 \le X < 30) = \mathrm{P}(20 \le X \le 29)$
Expressing this in terms of cumulative probabilities:
$\mathrm{P}(20 \le X \le 29) = \mathrm{P}(X \le 29) - \mathrm{P}(X \le 19)$
Using a graphing calculator to evaluate the binomial cumulative distribution function (binomcdf):
$\mathrm{P}(20 \le X \le 29) \approx 0.96010 - 0.01240$
$\mathrm{P}(20 \le X \le 29) = 0.94770 \approx 0.948 \text{ (to 3 s.f.)}$

1. Find the probability that both brothers cycle on a given day.

Let $K$ be the event that Kai cycles, and $L$ be the event that Leo cycles.
$P(K \cap L) = P(L \mid K) \times P(K) = 0.9 \times \frac{3}{4} = \frac{27}{40}$

2. Find the probability that neither brother cycles on a given day.

$P(K' \cap L') = 1 - P(K \cup L)$
$= 1 - [P(K) + P(L) - P(K \cap L)]$
$= 1 - \left(\frac{3}{4} + \frac{5}{7} - \frac{27}{40}\right) = \frac{59}{280}$

3. Calculate the probability of the specific arrangement of days over the 5 weeks.

The 5 weeks consist of $5 \times 7 = 35$ days.
The question specifies exactly 5 days of 'neither cycle' and the remaining $35 - 5 = 30$ days of 'both cycle'.
$P = \binom{35}{5} \times P(K' \cap L')^5 \times P(K \cap L)^{30}$
$= \binom{35}{5} \left(\frac{59}{280}\right)^5 \left(\frac{27}{40}\right)^{30}$
$\approx 0.0010198 \approx 0.00102 \text{ (3 s.f.)}$