The amusement park holds a game of chance in which entrance tickets to the amusement park can be won. At the beginning of the game, one rolls a die whose faces are numbered with the numbers 1 to 6. If one obtains the number 6, one may then spin a wheel of fortune with three sectors once (see schematic diagram). If sector $K$ is obtained, one wins a child ticket worth 28 euros; for sector $E$, an adult ticket worth 36 euros. For sector $N$, one wins nothing. The central angle of sector $N$ is $160 ^ { \circ }$. The sizes of sectors $K$ and $E$ are chosen such that the average winnings per game amount to three euros. Determine the size of the central angles of sectors $K$ and $E$.

21. If $\mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) \propto \mathrm { i }$, where $\mathrm { i } = 1,2,3 , \ldots \mathrm { n }$, then $\lim _ { \mathrm { n } \rightarrow \infty } \mathrm { P } ( \mathrm { w } )$ is equal to
(A) 1
(B) $\frac { 2 } { 3 }$
(C) $\frac { 3 } { 4 }$
(D) $\frac { 1 } { 4 }$
Sol. (B)
$$\begin{aligned} & \mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = \mathrm { ki } \\ & \Sigma \mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = 1 \\ & \Rightarrow \mathrm { k } = \frac { 2 } { \mathrm { n } ( \mathrm { n } + 1 ) } \\ & \lim _ { \mathrm { n } \rightarrow \infty } \mathrm { P } ( \mathrm { w } ) = \lim _ { \mathrm { n } \rightarrow \infty } \sum _ { \mathrm { i } = 1 } ^ { \mathrm { n } } \frac { 2 \mathrm { i } ^ { 2 } } { \mathrm { n } ( \mathrm { n } + 1 ) ^ { 2 } } = \lim _ { \mathrm { n } \rightarrow \infty } \frac { 2 \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { \mathrm { n } ( \mathrm { n } + 1 ) ^ { 2 } 6 } = \frac { 2 } { 3 } \end{aligned}$$

1. If $\mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = \mathrm { c }$, where c is a constant then $\mathrm { P } \left( \mathrm { u } _ { \mathrm { n } } / \mathrm { w } \right)$ is equal to
   (A) $\frac { 2 } { \mathrm { n } + 1 }$
   (B) $\frac { 1 } { n + 1 }$
   (C) $\frac { n } { n + 1 }$
   (D) $\frac { 1 } { 2 }$

Sol. (A)
$$\mathrm { P } \left( \frac { \mathrm { u } _ { \mathrm { n } } } { \mathrm { w } } \right) = \frac { \mathrm { c } \left( \frac { \mathrm { n } } { \mathrm { n } + 1 } \right) } { \mathrm { c } \left( \frac { \Sigma \mathrm { i } } { ( \mathrm { n } + 1 } \right) } = \frac { 2 } { \mathrm { n } + 1 }$$

1. If n is even and E denotes the event of choosing even numbered urn $\left( \mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = \frac { 1 } { \mathrm { n } } \right)$, then the value of $\mathrm { P } ( \mathrm { w } / \mathrm { E } )$ is
   (A) $\frac { n + 2 } { 2 n + 1 }$
   (B) $\frac { n + 2 } { 2 ( n + 1 ) }$
   (C) $\frac { n } { n + 1 }$
   (D) $\frac { 1 } { n + 1 }$

Sol. (B)

$$\mathrm { P } \left( \frac { \mathrm { w } } { \mathrm { E } } \right) = \frac { 2 + 4 + 6 + \cdots \mathrm { n } } { \frac { \mathrm { n } ( \mathrm { n } + 1 ) } { 2 } } = \frac { \mathrm { n } + 2 } { 2 ( \mathrm { n } + 1 ) }$$

Comprehension II

Suppose we define the definite integral using the following formula $\int _ { a } ^ { b } f ( x ) d x = \frac { b - a } { 2 } ( f ( a ) + f ( b ) )$, for more accurate result for $\mathrm { c } \in ( \mathrm { a } , \mathrm { b } ) \mathrm { F } ( \mathrm { c } ) = \frac { \mathrm { c } - \mathrm { a } } { 2 } ( \mathrm { f } ( \mathrm { a } ) + \mathrm { f } ( \mathrm { c } ) ) + \frac { \mathrm { b } - \mathrm { c } } { 2 } ( \mathrm { f } ( \mathrm { b } ) + \mathrm { f } ( \mathrm { c } ) ) \quad$. When $\mathrm { c } = \frac { \mathrm { a } + \mathrm { b } } { 2 } , \int _ { \mathrm { a } } ^ { \mathrm { b } } \mathrm { f } ( \mathrm { x } ) \mathrm { dx } = \frac { \mathrm { b } - \mathrm { a } } { 4 } ( \mathrm { f } ( \mathrm { a } ) + \mathrm { f } ( \mathrm { b } ) + 2 \mathrm { f } ( \mathrm { c } ) )$.

Let $S = \left\{ w _ { 1 } , w _ { 2 } , \ldots \right\}$ be the sample space associated to a random experiment. Let $P \left( w _ { n } \right) = \frac { P \left( w _ { n - 1 } \right) } { 2 } , n \geq 2$ . Let $A = \{ 2 k + 3 l ; k , l \in \mathbb { N } \}$ and $B = \left\{ w _ { n } ; n \in A \right\}$. Then $P ( B )$ is equal to (1) $\frac { 3 } { 32 }$ (2) $\frac { 3 } { 64 }$ (3) $\frac { 1 } { 16 }$ (4) $\frac { 1 } { 32 }$

There are nine cards on which the integers from 1 to 9 are written in a box. Two cards are taken simultaneously from this box. Let $S$ denote the sum of the numbers written on the two cards.
(1) The probability that $S$ is 5 or less is $\frac { \mathbf { A } } { \mathbf { B } }$. Let us assign a score to the result $S$.
When $S$ is 5 or less the score is $10 - S$, and when it is greater than 5 the score is 2. Then the expected value of the score is $\frac { \mathbf { C D } } { \mathbf { E F } }$.
(2) Let us perform the above trial twice, returning the two cards to the box before the second trial.
(i) The probability that $S$ is 5 or less in both trials is $\frac { \mathbf { G } } { \mathbf { H } }$.
(ii) The probability that $S$ is 5 or less in at least one trial is $\frac { \mathbf { J K } } { \mathbf { L M } }$.

Q2 Consider dice-X with the following property: when it is rolled, for numbers 1 through 5 the probabilities of that number's coming up are all the same, but the probability that 6 comes up is twice that of any other number.
(1) Denote by $p$ the probability that a particular number 1 through 5 comes up. The probability that number 6 comes up is $\mathbf { L }$ p. Since the probability of the whole event is $\mathbf { M }$, we have $p = \frac { \mathbf { N } } { \mathbf { O } }$.
(2) Dice-X is rolled twice in succession. Let us denote by $A$ the event that for both rolls the number that comes up is 1 through 5, and by $B$ the event that number 6 comes up at least once. Then the probability of event $A$, $P(A)$, and that of event $B$, $P(B)$, are
$$P ( A ) = \frac { \mathbf { PQ } } { \mathbf { RS } } , \quad P ( B ) = \frac { \mathbf { TU } } { \mathbf { VW } } .$$
Hence, we see that $\mathbf { X }$. (For $\mathbf { X }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( A )$ is less than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(1) $P ( A )$ is less than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(2) $P ( A )$ and $P ( B )$ are the same.
(3) $P ( A )$ is greater than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(4) $P ( A )$ is greater than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(3) Next, dice-X is rolled three times in succession. Let us denote by $C$ the event that for all three rolls the number which comes up is 1 through 5, and by $D$ the event that the number 6 comes up at least once. When the probability $P(C)$ is compared with the probability $P(D)$, we see that $\mathbf { Y }$. (For $\mathbf { Y }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is not less than twice $P ( C )$.
(1) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is less than twice $P ( C )$.
(2) $P ( C )$ and $P ( D )$ are the same.
(3) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is not less than twice $P ( D )$.
(4) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is less than twice $P ( D )$.

(Course 2) Q2 Consider dice-X with the following property: when it is rolled, for numbers 1 through 5 the probabilities of that number's coming up are all the same, but the probability that 6 comes up is twice that of any other number.
(1) Denote by $p$ the probability that a particular number 1 through 5 comes up. The probability that number 6 comes up is $\mathbf { L }$ p. Since the probability of the whole event is $\mathbf { M }$, we have $p = \frac { \mathbf { N } } { \mathbf { O } }$.
(2) Dice-X is rolled twice in succession. Let us denote by $A$ the event that for both rolls the number that comes up is 1 through 5, and by $B$ the event that number 6 comes up at least once. Then the probability of event $A$, $P(A)$, and that of event $B$, $P(B)$, are
$$P ( A ) = \frac { \mathbf { PQ } } { \mathbf { RS } } , \quad P ( B ) = \frac { \mathbf { TU } } { \mathbf { VW } }$$
Hence, we see that $\mathbf { X }$. (For $\mathbf { X }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( A )$ is less than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(1) $P ( A )$ is less than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(2) $P ( A )$ and $P ( B )$ are the same.
(3) $P ( A )$ is greater than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(4) $P ( A )$ is greater than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(3) Next, dice-X is rolled three times in succession. Let us denote by $C$ the event that for all three rolls the number which comes up is 1 through 5, and by $D$ the event that the number 6 comes up at least once. When the probability $P(C)$ is compared with the probability $P(D)$, we see that $\mathbf { Y }$. (For $\mathbf { Y }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is not less than twice $P ( C )$.
(1) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is less than twice $P ( C )$.
(2) $P ( C )$ and $P ( D )$ are the same.
(3) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is not less than twice $P ( D )$.
(4) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is less than twice $P ( D )$.