Problem List

Assigned by Dr. Le Vi

Problem 1.

Three fair and identical dice are rolled. Calculate the probability for the following cases:

• A. The sum of the results is 7.
• B. The number of dots on each die is different across all three dice.
• C. The product of the results is divisible by 3.
• D. The product of the results is divisible by 6.

Solution 1.

Total number of outcomes: $ 6^3 = 216 $.

A. The sum of the results is 7.

Approach 1 - Counting: We can manually count each permutation, we can also reason how many are there:

Partition	Frequency Count
$(1,1,5)$	$3! / (2!1!) = 3$
$(1,2,4)$	$3! / (1!1!1!) = 6$
$(1,3,3)$	$3! / (1!2!) = 3$
$(2,2,3)$	$3! / (2!1!) = 3$

So there are $15$ valid cases, therefore we can calculate the result: $\frac{15}{216} = \frac{5}{72}$.

Approach 2 - Stars & Bars: We can also reason that there are $15$ valid cases through the Stars & Bars method:

Step 1: Define the Problem

We need to find the number of integer solutions to:

\[x_1 + x_2 + x_3 = 7\]

Where each $x_i$ must satisfy:

\[1 \leq x_i \leq 6\]

Step 2: Change of Variables

Since each die must be at least 1, define new variables:

\[y_i = x_i - 1 \implies 0 \leq y_i \leq 5\]

Rewriting the equation gives us:

\[(y_1 + 1) + (y_2 + 1) + (y_3 + 1) = 7 \\[10pt] \leftrightarrow y_1 + y_2 + y_3 = 4 (0 \leq y_i \leq 5)\]

Step 3: Apply Stars & Bars

\[\binom{4+3-1}{3-1} = \binom{6}{2} = 15\]

So there are $15$ valid cases, therefore we can calculate the result: $\frac{15}{216} = \frac{5}{72}$.

B. The number of dots on each die is different across all three dice.

The number of ordered triples with all distinct numbers is $ 6 \times 5 \times 4 = 120 $.

The probability is $ \frac{120}{216} = \frac{5}{9} $.

C. The product of the results is divisible by 3.

For the product of the three dice to be divisible by 3, at least one of the dice must show a multiple of 3. The multiples of 3 on a die are 3 and 6.

Instead of directly calculating the probability that the product is divisible by 3, we calculate the probability that the product is not divisible by 3. This happens only if none of the dice show a multiple of 3.

The complementary probability (product not divisible by 3) occurs when all numbers are 1, 2, 4, or 5. The number of outcomes where none of the dice show a multiple of 3 is:

\[4 \times 4 \times 4 = 64\]

The probability that the product is divisible by 3 is the complement of the above:

\[1 - \frac{64}{216} = \frac{152}{216} = \frac {19}{27}\]

D. The product of the results is divisible by 6.

For the product of the three dice to be divisible by 6, it must be divisible by both 2 and 3. This means:

• At least one die must show a multiple of 2 (even number: 2, 4, 6).
• At least one die must show a multiple of 3 (3, 6).

We calculate the probability that the product is not divisible by 6. Using the inclusion-exclusion principle, the probability of the complement is:

\[P(\nmid 2)+P(\nmid 3)−P(\nmid 2 \space \vee \nmid 3).\]

Probability of not divisible by 2: This happens only if all dice show odd numbers (1, 3, 5). The number of such outcomes is:

\[3 \times 3 \times 3 = 27\]

Probability of not divisible by 3: This happens only if none of the dice show a multiple of 3 (1, 2, 4, 5). The number of such outcomes is:

\[4 \times 4 \times 4 = 64\]

Probability of not divisible by 2 and not divisible by 3: This happens only if all dice show numbers that are both odd and not multiples of 3. The allowed numbers are 1 and 5. The number of such outcomes is:

\[2 \times 2 \times 2 = 8\]

The probability that the product of the results is divisible by 6 is:

\[1 - \left( \frac{27}{216} + \frac{64}{216} - \frac{8}{216} \right) = \frac{133}{216}\]

Problem 2.

From a group of 10 people, consisting of 6 men and 4 women, 7 people are selected. Calculate the probability that:

• A. The selected group contains exactly 4 men.
• B. The selected group contains at least 2 women.

Solution 2.

A. The selected group contains exactly 4 men.

Total ways to choose 7 people from 10:

\[\binom{10}{7} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\]

Favorable outcomes (4 men and 3 women):

• Choose 4 men from 6: $ \binom{6}{4} = 15 $.
• Choose 3 women from 4: $ \binom{4}{3} = 4 $.
• Total favorable outcomes: $ 15 \times 4 = 60 $.

Probability: $ \frac{60}{120} = \frac{1}{2} $.

B. The selected group contains at least 2 women

Complementary probability (fewer than 2 women):

• 1 woman and 6 men:

\[\binom{4}{1} \times \binom{6}{6} = 4 \times 1 = 4\]

• Total complementary outcomes: $ 0 + 4 = 4 $.
• Complementary probability: $ \frac{4}{120} = \frac{1}{30} $.

Probability of at least 2 women: $ 1 - \frac{1}{30} = \frac{29}{30} $.

Problem 3.

There are 12 books, including 3 books by the author “Nguyen Nhat Anh”. Arrange all 12 books on a bookshelf. Calculate the probability that no two books by “Nguyen Nhat Anh” are placed next to each other.

Solution 3.

Total Arrangements: Assuming the 9 untitled books (lets call them red books) and the 3 by Nguyen Nhat Anh (lets call them blue books) are indistinguishable respectively, the total number of arrangements is:

\[\binom{12}{3} = 220\]

Favorable Arrangements (No Two Blue Books Together): First, place the 9 red books in a row. This creates 10 gaps (one before the first red book, one after the last red book, and 8 between the red books). To ensure that no two blue books are adjacent, each blue book must be placed in a different gap. Therefore, the number of favorable arrangements is:

\[\binom{10}{3} = 120\]

Probability Calculation: The probability that no two blue books are next to each other is the ratio of favorable arrangements to the total arrangements:

\[\frac{120}{220} = \frac{6}{11}\]

Problem 4.

Binh, An, Chien, and Thang share 20 candies among themselves. Calculate the probability that Binh has at least 3 candies and An has at least 4 candies.

Solution 4.

Let’s denote the number of candies that Binh, An, Chien, and Thang receive by $ x_B $, $ x_A $, $ x_C $ and $ x_T $ respectively. Since all 20 candies are distributed, we have:

\[x_B + x_A + x_C + x_T = 20\]

Total Number of Distributions: Without any restrictions, the number of nonnegative integer solutions is given by the stars and bars formula:

\[\binom{20 + 4 - 1}{4 - 1} = \binom{23}{3} = 1771\]

Favorable Distributions: We are given that Binh must have at least 3 candies and An at least 4. Define new variables:

\[y_B = x_B - 3 \\[10pt] y_A = x_A - 4\]

Substitute these into the total and simplifies:

\[(y_B + 3) + (y_A + 4) + x_C + x_T = 20 \\[10pt] \leftrightarrow y_B + y_A + x_C + x_T = 13\]

The number of nonnegative integer solutions is:

\[\binom{13 + 4 - 1}{4 - 1} = \binom{16}{3} = 560\]

Probability Calculation: The probability that the conditions are met is the ratio of the favorable distributions to the total distributions:

\[\frac{560}{1771} = \frac{80}{253}\]