A customer is chosen at random from those who bought a melon during week 1. Among customers who buy a melon in a given week, $90\%$ of them buy a melon the following week; among customers who do not buy a melon in a given week, $60\%$ of them do not buy a melon the following week. For $n \geqslant 1$, we set $p_n = P(A_n)$, where $A_n$ is the event ``the customer buys a melon during week $n$''. Thus $p_1 = 1$. Prove that, for all integer $n \geqslant 1$: $p_{n+1} = 0{,}5\, p_n + 0{,}4$.

21. Solution: The possible values of $X$ are $-1, 0, 1$.
$$\begin{aligned} & P(X = -1) = (1-\alpha)\beta, \\ & P(X = 0) = \alpha\beta + (1-\alpha)(1-\beta), \\ & P(X = 1) = \alpha(1-\beta), \end{aligned}$$
Therefore the distribution table of $X$ is

$X$	$-1$	$0$	$1$
$P$	$(1-\alpha)\beta$	$\alpha\beta + (1-\alpha)(1-\beta)$	$\alpha(1-\beta)$

(2) (i) From (1) we have $a = 0.4$, $b = 0.5$, $c = 0.1$. Therefore $p_i = 0.4p_{i-1} + 0.5p_i + 0.1p_{i+1}$, so $0.1(p_{i+1} - p_i) = 0.4(p_i - p_{i-1})$, that is $p_{i+1} - p_i = 4(p_i - p_{i-1})$. Since $p_1 - p_0 = p_1 \neq 0$, the sequence $\{p_{i+1} - p_i\}$ $(i = 0, 1, 2, \cdots, 7)$ is a geometric sequence with common ratio 4 and first term $p_1$.
(ii) From (i) we obtain $p_8 = p_8 - p_7 + p_7 - p_6 + \cdots + p_1 - p_0 + p_0 = (p_8 - p_7) + (p_7 - p_6) + \cdots + (p_1 - p_0) = \frac{4^8 - 1}{3}p_1$. Since $p_8 = 1$, we have $p_1 = \frac{3}{4^8 - 1}$, therefore $p_4 = (p_4 - p_3) + (p_3 - p_2) + (p_2 - p_1) + (p_1 - p_0) = \frac{4^4 - 1}{3}p_1 = \frac{1}{257}$. $p_4$ represents the probability of ultimately concluding that drug A is more effective. From the calculation result, we can see that when drug A has a cure rate of 0.5 and drug B has a cure rate of 0.8, the probability of concluding that drug A is more effective is $p_4 = \frac{1}{257} \approx 0.0039$. The probability of reaching an incorrect conclusion is very small, indicating that this experimental scheme is reasonable.

Let $\left(X_{n}\right)_{n \in \mathbb{N}}$ be a sequence of mutually independent Rademacher random variables and $S_{n} = \sum_{j=1}^{n} X_{j}$. For every $n \in \mathbb{N}^{*}$ and every $k \in \mathbb{Z}$, by setting $p_{n}(k) = \mathbb{P}\left(S_{n} = k\right)$, show that $$\frac{p_{n+1}(k) - p_{n}(k)}{\tau} = \frac{\delta^{2}}{2\tau} \frac{p_{n}(k+1) - 2p_{n}(k) + p_{n}(k-1)}{\delta^{2}}$$

We have an infinite supply of black and white balls. An urn initially contains one black ball and one white ball. We perform a sequence of draws according to the following protocol:

• we randomly draw a ball from the urn;
• we replace the drawn ball in the urn;
• we add to the urn a ball of the same color as the drawn ball.

We define the sequence $(X_{n})_{n \in \mathbb{N}}$ of random variables by $X_{0} = 1$ and, for all integers $n \geqslant 1$, $X_{n}$ gives the number of white balls in the urn after $n$ draws.
Let $n$ and $k$ be two integers greater than or equal to 1. Establish that $$P(X_{n} = k) = \frac{k-1}{n+1} P(X_{n-1} = k-1) + \frac{n+1-k}{n+1} P(X_{n-1} = k).$$

Let $n \in \mathbb{N}^{*}$. Show that $$1 = \sum _ { k = 0 } ^ { n } P \left( S _ { k } = 0 _ { d } \right) P ( R > n - k )$$

Let $T$ be the random variable, defined on $\Omega$ and taking values in $\mathbb { N }$, equal to the first instant when the particle returns to the origin, if this instant exists, and equal to 0 if the particle never returns to the origin: $$\forall \omega \in \Omega , \quad T ( \omega ) = \begin{cases} 0 & \text { if } \forall k \in \mathbb { N } ^ { * } , S _ { k } ( \omega ) \neq 0 \\ \min \left\{ k \in \mathbb { N } ^ { * } \mid S _ { k } ( \omega ) = 0 \right\} & \text { otherwise } \end{cases}$$ Show that $T$ takes even values and that, for all $n \in \mathbb { N } , \mathbb { P } ( T = 2 n + 2 ) = 2 C _ { n } p ^ { n + 1 } ( 1 - p ) ^ { n + 1 }$.

9. (a) A coin has probability p of showing head when tossed. It is tossed n times. Let pn , denote the probability that no two (or more) consecutive heads occur. Prove that $\mathrm { p } 1 = 1 , \mathrm { p } ^ { 2 } = 1 - \mathrm { p } _ { 2 }$ and $\mathrm { p } = ( 1 -$ p ). $\mathrm { pn } - 1 + \mathrm { p } ( 1 - \mathrm { p } ) \mathrm { pn } - 2$ for all $\mathrm { n } \geq 3$.
(b) In (a), prove by induction on n , that $\mathrm { pn } = \mathrm { A } \alpha \mathrm { n } + \mathrm { B } \beta \mathrm { n }$ for all $\mathrm { n } \geq 1$, where $\alpha$ and $\beta$ are the roots of the quadratic x2-(1-p) $x - p ( 1 - p ) = 0$ and
$$A = \frac { p ^ { 2 } + \beta - 1 } { \alpha \beta - \alpha ^ { 2 } } , B = \frac { p ^ { 2 } + \alpha - 1 } { \alpha \beta - \beta ^ { 2 } } .$$

1. Let ABC and PQR be any two triangles in the same plane. Assume that the perpendiculars from the points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ to the sides $\mathrm { QR } , \mathrm { RP } , \mathrm { PQ }$ respectively are concurrent. Using vector methods or otherwise, prove that the perpendiculars from $\mathrm { P } , \mathrm { Q } , \mathrm { R }$ to $\mathrm { BC } , \mathrm { CA } , \mathrm { AB }$ respectively are also concurrent.

In a game two players $A$ and $B$ take turns in throwing a pair of fair dice starting with player $A$ and total of scores on the two dice, in each throw is noted. $A$ wins the game if he throws a total of 6 before $B$ throws a total of 7 and $B$ wins the game if he throws a total of 7 before $A$ throws a total of six. The game stops as soon as either of the players wins. The probability of $A$ winning the game is:
(1) $\frac { 5 } { 31 }$
(2) $\frac { 31 } { 61 }$
(3) $\frac { 5 } { 6 }$
(4) $\frac { 30 } { 61 }$

Problem 6

Consider the following procedure that generates a sequence of random variables that take the value 0 or 1. For an integer $n \geq 1$, we denote the $n$-th random variable of a sequence generated by the procedure as $X _ { n }$.

• $X _ { 1 }$ becomes $X _ { 1 } = 0$ with a probability $\frac { 2 } { 3 }$ and $X _ { 1 } = 1$ with a probability $\frac { 1 } { 3 }$.
• For an integer $n = 1,2 , \ldots$ in order, the following is repeated until the procedure terminates:
• The procedure terminates with a probability $p ( 0 < p < 1 )$ if $X _ { n } = 0$ and with a probability $q ( 0 < q < 1 )$ if $X _ { n } = 1$. Here, $p$ and $q$ are constants.
• If the procedure does not terminate as above, $X _ { n + 1 }$ becomes $X _ { n + 1 } = 0$ with a probability $\frac { 2 } { 3 }$ and becomes $X _ { n + 1 } = 1$ with a probability $\frac { 1 } { 3 }$.

When the procedure terminates at $n = \ell$, a sequence of length $\ell$, composed of random variables $\left( X _ { 1 } , \ldots , X _ { \ell } \right)$, is generated, and no further random variables are generated. Answer the following questions.
I. For an integer $k \geq 1$, consider the following matrix:
$$\boldsymbol { P } _ { k } = \left( \begin{array} { c c } \operatorname { Pr } \left( X _ { n + k } = 0 \mid X _ { n } = 0 \right) & \operatorname { Pr } \left( X _ { n + k } = 1 \mid X _ { n } = 0 \right) \\ \operatorname { Pr } \left( X _ { n + k } = 0 \mid X _ { n } = 1 \right) & \operatorname { Pr } \left( X _ { n + k } = 1 \mid X _ { n } = 1 \right) \end{array} \right)$$
Here, $\operatorname { Pr } ( A \mid B )$ is the conditional probability of an event $A$ given that an event $B$ has occurred.

1. Express $\boldsymbol { P } _ { 1 }$ and $\boldsymbol { P } _ { 2 }$ using $p$ and $q$.
2. Express $\boldsymbol { P } _ { 3 }$ using $\boldsymbol { P } _ { 1 }$.
3. $\boldsymbol { P } _ { k }$ can be expressed as $\boldsymbol { P } _ { k } = \gamma _ { k } \boldsymbol { P } _ { 1 }$ using a real number $\gamma _ { k }$. Find $\gamma _ { k }$.

II. For an integer $m \geq 2$, find the respective probabilities that $X _ { m } = 0$ and $X _ { m } = 1$, given that the procedure does not terminate before $n = m$.
III. Find the expected value and the variance of the length of the sequence, $\ell$, generated by the procedure. If necessary, you may use $\sum _ { m = 1 } ^ { \infty } m r ^ { m - 1 } = \frac { 1 } { ( 1 - r ) ^ { 2 } }$ and $\sum _ { m = 1 } ^ { \infty } m ^ { 2 } r ^ { m - 1 } = \frac { 1 + r } { ( 1 - r ) ^ { 3 } }$ for a real number $r$ whose absolute value is smaller than 1.
IV. For an integer $k \geq 1$, find the probability $\operatorname { Pr } \left( X _ { n } = 0 \mid X _ { n + k } = 1 \right)$.