For two events $A , B$, $$\mathrm { P } ( A ) = \frac { 2 } { 5 } , \quad \mathrm { P } ( B \mid A ) = \frac { 5 } { 6 }$$ what is the value of $\mathrm { P } ( A \cap B )$? [3 points]
(1) $\frac { 1 } { 3 }$
(2) $\frac { 4 } { 15 }$
(3) $\frac { 1 } { 5 }$
(4) $\frac { 2 } { 15 }$
(5) $\frac { 1 } { 15 }$

A company has a total of 60 employees, and each employee belongs to one of two departments, A or B. Department A has 20 employees and Department B has 40 employees. 50\% of the employees in Department A are women. 60\% of the women employees in the company belong to Department B. When one employee is randomly selected from the 60 employees and is found to belong to Department B, the probability that this employee is a woman is $p$. Find the value of $80p$. [4 points]

A survey was conducted on 500 students at a high school regarding their desire to visit regions A and B for cultural exploration. The results are as follows. (Unit: students)

Region B	Wish	Do not wish	Total
Wish	140	310	450
Do not wish	40	10	50
Total	180	320	500

When one student is randomly selected from this high school and is found to wish to visit region A, what is the probability that this student also wishes to visit region B? [3 points]
(1) $\frac { 19 } { 45 }$
(2) $\frac { 23 } { 45 }$
(3) $\frac { 3 } { 5 }$
(4) $\frac { 31 } { 45 }$
(5) $\frac { 7 } { 9 }$

A die is rolled twice. Given that the number 6 does not appear at all, what is the probability that the sum of the two numbers is a multiple of 4? [3 points]
(1) $\frac { 4 } { 25 }$
(2) $\frac { 1 } { 5 }$
(3) $\frac { 6 } { 25 }$
(4) $\frac { 7 } { 25 }$
(5) $\frac { 8 } { 25 }$

For two events $A , B$, $A$ and $B ^ { C }$ are mutually exclusive events, and $$\mathrm { P } ( A ) = \frac { 1 } { 3 } , \mathrm { P } \left( A ^ { C } \cap B \right) = \frac { 1 } { 6 }$$ What is the value of $\mathrm { P } ( B )$? (Here, $A ^ { C }$ is the complement of $A$.) [3 points]
(1) $\frac { 5 } { 12 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 7 } { 12 }$
(4) $\frac { 2 } { 3 }$
(5) $\frac { 3 } { 4 }$

A survey was conducted on 200 students at a school regarding their preferences for experiential activities. The students who participated in this survey chose one of cultural experience or ecological research, and the number of students who chose each activity is as follows.

Classification	Cultural Experience	Ecological Research	Total
Male Students	40	60	100
Female Students	50	50	100
Total	90	110	200

When one student is randomly selected from the 200 students who participated in this survey and is a student who chose ecological research, what is the probability that this student is a female student? [3 points]
(1) $\frac { 5 } { 11 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 6 } { 11 }$
(4) $\frac { 5 } { 9 }$
(5) $\frac { 3 } { 5 }$

For two events $A$ and $B$, $$\mathrm { P } ( B \mid A ) = \frac { 1 } { 4 } , \quad \mathrm { P } ( A \mid B ) = \frac { 1 } { 3 } , \quad \mathrm { P } ( A ) + \mathrm { P } ( B ) = \frac { 7 } { 10 }$$ what is the value of $\mathrm { P } ( A \cap B )$? [3 points]
(1) $\frac { 1 } { 7 }$
(2) $\frac { 1 } { 8 }$
(3) $\frac { 1 } { 9 }$
(4) $\frac { 1 } { 10 }$
(5) $\frac { 1 } { 11 }$

There is a basket containing at least 10 white balls and at least 10 black balls, and an empty bag. Using one die, the following trial is performed.
Roll the die once. If the result is 5 or more, put 2 white balls from the basket into the bag. If the result is 4 or less, put 1 black ball from the basket into the bag.
When the above trial is repeated 5 times, let $a _ { n }$ and $b _ { n }$ denote the number of white balls and black balls in the bag after the $n$-th trial ($1 \leq n \leq 5$) respectively. Given that $a _ { 5 } + b _ { 5 } \geq 7$, what is the probability that there exists a natural number $k$ ($1 \leq k \leq 5$) such that $a _ { k } = b _ { k }$? If this probability is $\frac { q } { p }$, find the value of $p + q$. (Here, $p$ and $q$ are coprime natural numbers.) [4 points]

A bag contains 1 white ball marked with 1, 1 white ball marked with 2, 1 black ball marked with 1, and 3 black balls marked with 2. We perform a trial of simultaneously drawing 3 balls from the bag. Let $A$ be the event that among the 3 balls drawn, 1 is white and 2 are black, and let $B$ be the event that the product of the numbers on the 3 balls is 8. What is the value of $\mathrm { P } ( A \cup B )$? [3 points]
(1) $\frac { 11 } { 20 }$
(2) $\frac { 3 } { 5 }$
(3) $\frac { 13 } { 20 }$
(4) $\frac { 7 } { 10 }$
(5) $\frac { 3 } { 4 }$

There are 6 cards with natural numbers 1 through 6 written on the front and 0 written on the back. These 6 cards are placed so that the natural number $k$ is visible in the $k$-th position for natural numbers $k$ not exceeding 6.
Using these 6 cards and one die, we perform the following trial:
Roll the die once. If the result is $k$, flip the card in the $k$-th position and place it back in its original position.
After repeating this trial 3 times, given that the sum of all numbers visible on the 6 cards is even, what is the probability that the die shows 1 exactly once? The probability is $\frac { q } { p }$. Find the value of $p + q$. (Here, $p$ and $q$ are coprime natural numbers.) [4 points]

There is one bag and two boxes A and B. The bag contains 4 cards with the numbers $1, 2, 3, 4$ written on them, one number per card. Box A contains more than 8 white balls and more than 8 black balls, and box B is empty. Using this bag and the two boxes A and B, the following trial is performed.
A card is randomly drawn from the bag, the number on the card is confirmed, and the card is returned to the bag.
If the confirmed number is 1, 1 white ball from box A is placed into box B. If the confirmed number is 2 or 3, 1 white ball and 1 black ball from box A are placed into box B. If the confirmed number is 4, 2 white balls and 1 black ball from box A are placed into box B.
After repeating this trial 4 times, given that the total number of balls in box B is 8, find the probability that the number of black balls in box B is 2. [4 points]
(1) $\frac{3}{70}$
(2) $\frac{2}{35}$
(3) $\frac{1}{14}$
(4) $\frac{3}{35}$
(5) $\frac{1}{10}$

For two events $A$ and $B$, $$\mathrm{P}(A \mid B) = \mathrm{P}(A) = \frac{1}{2}, \quad \mathrm{P}(A \cap B) = \frac{1}{5}$$ What is the value of $\mathrm{P}(A \cup B)$? [3 points]
(1) $\frac{1}{2}$
(2) $\frac{3}{5}$
(3) $\frac{7}{10}$
(4) $\frac{4}{5}$
(5) $\frac{9}{10}$

For two events $A , B$, $$\mathrm { P } ( A ) = \frac { 2 } { 5 } , \quad \mathrm { P } ( B \mid A ) = \frac { 1 } { 4 } , \quad \mathrm { P } ( A \cup B ) = 1$$ What is the value of $\mathrm { P } ( B )$? [3 points]
(1) $\frac { 7 } { 10 }$
(2) $\frac { 3 } { 4 }$
(3) $\frac { 4 } { 5 }$
(4) $\frac { 17 } { 20 }$
(5) $\frac { 9 } { 10 }$

There are 16 balls and six empty boxes with the natural numbers 1 through 6 written on them. A trial is performed using one die.
When the die is rolled and the result is $k$: If $k$ is odd, place 1 ball each in the boxes labeled $1, 3, 5$, and if $k$ is even, place 1 ball each in the boxes labeled with the divisors of $k$.
After repeating this trial 4 times, given that the sum of all balls in the six boxes is odd, what is the probability that the number of balls in the box labeled 3 is 1 more than the number of balls in the box labeled 2? [4 points]
(1) $\frac { 1 } { 8 }$
(2) $\frac { 3 } { 16 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 5 } { 16 }$
(5) $\frac { 3 } { 8 }$

A chess player plays one game each against three chess players A, B, and C, with the results of each game being independent. The probabilities that the player wins against A, B, and C are $p_1, p_2, p_3$ respectively, where $p_3 > p_2 > p_1 > 0$. Let $p$ denote the probability that the player wins two consecutive games. Then
A. $p$ is independent of the order of games against A, B, and C
B. $p$ is maximum when the player plays against A in the second game
C. $p$ is maximum when the player plays against B in the second game
D. $p$ is maximum when the player plays against C in the second game

50 people registered for the soccer club, 60 people registered for the table tennis club, and 70 people registered for either the soccer or table tennis club. If a person is known to have registered for the soccer club, the probability that they also registered for the table tennis club is
A. $0.8$
B. $0.4$
C. $0.2$
D. $0.1$

We fix $n \in \mathbf { N } ^ { * }$ and draw successively and with replacement two integers $p$ and $q$ according to a uniform distribution on $\llbracket 1 , n \rrbracket$. Using the result $H_n \sim \ln n$ as $n \to +\infty$, show that $$\mathbf { P } \left( A _ { n } \cup B _ { n } \right) \sim \frac { \ln n } { n } \quad ( n \rightarrow + \infty )$$

Let $( s , i , r ) \in E$ where $E = \{ ( s , i , r ) \in \mathbf{N}^3,\, s + i + r = M \}$. Conditional on the event $\left( \left( \tilde { S } _ { n } , \tilde { I } _ { n } , \tilde { R } _ { n } \right) = ( s , i , r ) \right)$, what is the probability, denoted $p ( i )$, for a susceptible person to be infected during this day?
Each of the $s$ healthy persons meets, independently of the others, $K$ persons chosen at random from the $M$ persons in the total population. As soon as at least one of the meetings is with an infected person, the healthy person becomes infected the next morning.

Let $Z$ be a random variable taking values in $\{ 0 , \ldots , M \}$. Show that:
$$\mathbf { E } [ Z ] = \sum _ { ( s , i , r ) \in E } \left( \sum _ { k = 0 } ^ { M } k \mathbf { P } \left( Z = k \mid \left( \tilde { S } _ { n } , \tilde { I } _ { n } , \tilde { R } _ { n } \right) = ( s , i , r ) \right) \right) \mathbf { P } \left( \left( \tilde { S } _ { n } , \tilde { I } _ { n } , \tilde { R } _ { n } \right) = ( s , i , r ) \right).$$

A bag contains some candies, $\frac { 2 } { 5 }$ of them are made of white chocolate and the remaining $\frac { 3 } { 5 }$ are made of dark chocolate. Out of the white chocolate candies, $\frac { 1 } { 3 }$ are wrapped in red paper, the rest are wrapped in blue paper. Out of the dark chocolate candies, $\frac { 2 } { 3 }$ are wrapped in red paper, the rest are wrapped in blue paper. If a randomly selected candy from the bag is found to be wrapped in red paper, then what is the probability that it is made up of dark chocolate?
(A) $\frac { 2 } { 3 }$
(B) $\frac { 3 } { 4 }$
(C) $\frac { 3 } { 5 }$
(D) $\frac { 1 } { 4 }$

A brand called Jogger's Pride produces pairs of shoes in three different units that are named $U _ { 1 } , U _ { 2 }$ and $U _ { 3 }$. These units produce $10 \% , 30 \% , 60 \%$ of the total output of the brand with the chance that a pair of shoes being defective is $20 \% , 40 \% , 10 \%$ respectively. If a randomly selected pair of shoes from the combined output is found to be defective, then what is the chance that the pair was manufactured in the unit $U _ { 3 }$?
(A) $30 \%$
(B) $15 \%$
(C) $\frac { 3 } { 5 } \times 100 \%$
(D) Cannot be determined from the given data.

Shubhaangi thinks she may be allergic to Bengal gram and takes a test that is known to give the following results:

• For people who really do have the allergy, the test says ``Yes'' $90 \%$ of the time.
• For people who do not have the allergy, the test says ``Yes'' $15 \%$ of the time.

If $2 \%$ of the population has the allergy and Shubhaangi's test says ``Yes'', then the chances that Shubhaangi does really have the allergy are
(A) $1 / 9$
(B) $6 / 55$
(C) $1 / 11$
(D) cannot be determined from the given data.

One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{2}{5}$
(D) $\frac{1}{5}$

Let $E^c$ denote the complement of an event $E$. Let $E$, $F$, $G$ be pairwise independent events with $P(G) > 0$ and $P(E \cap F \cap G) = 0$. Then $P(E^c \cap F^c | G)$ equals
(A) $P(E^c) + P(F^c)$
(B) $P(E^c) - P(F^c)$
(C) $P(E^c) - P(F)$
(D) $P(E) - P(F^c)$

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required.
The conditional probability that $X \geq 6$ given $X > 3$ equals
(A) $\frac { 125 } { 216 }$
(B) $\frac { 25 } { 216 }$
(C) $\frac { 5 } { 36 }$
(D) $\frac { 25 } { 36 }$