Computing Probabilities Using Combinatorics

Statement. Two urns contain balls that differ only in color. The first urn has two white and three black balls, and the second has three white and two black balls. One ball is drawn at random from each urn. What is the probability that at least one of the two balls is white?

Solution. Each urn contains 5 balls, so $5 \cdot 5 = 25$ pairs can be formed. Since the balls are numbered, all 25 pairs are distinct. Because the balls are drawn at random, this experiment has 25 equally likely outcomes.

Since the first urn contains 3 black balls and the second contains 2 black balls, there are $3 \cdot 2 = 6$ all-black pairs. Therefore, the number of outcomes favorable to the event “at least one ball is white” is $25 - 6 = 19$.

Answer: $\dfrac{19}{25}$.

Statement. Four dice are rolled. Find the probability that: 1) exactly one six appears (event $A$); 2) four different numbers appear (event $B$); 3) no six appears (event $C$); 4) at least one six appears (event $D$).

Solution. Number the dice 1 through 4. Any result is recorded as an ordered tuple $(a; b; c; d)$ where $a, b, c, d$ are the numbers showing on dice 1, 2, 3, and 4 respectively.

The total number of outcomes is $6^4$. All outcomes are equally likely.

1. The single six can appear on any of the four positions. On the other three positions, digits 1 through 5 can appear. The number of favorable outcomes is $4 \cdot 5^3$. Therefore $P(A) = \dfrac{4 \cdot 5^3}{6^4}$.

2. Four different digits form a 4-element ordered subset of $\{1, 2, 3, 4, 5, 6\}$. The number of favorable outcomes is $A_6^4$. Therefore $P(B) = \dfrac{A_6^4}{6^4} = \dfrac{6 \cdot 5 \cdot 4 \cdot 3}{6^4}$.

3. Each of the four positions can show any digit from 1 to 5. The number of favorable outcomes is $5^4$. Therefore $P(C) = \dfrac{5^4}{6^4}$.

4. The number of outcomes containing at least one six is $6^4 - 5^4$. Therefore

Statement. An inspector randomly selects 5 items from a batch of 20 for testing. If none of the selected items are defective, the entire batch is accepted. What is the probability that the inspector accepts a batch containing 7 defective items?

Solution. Since the inspector selects 5 items at random from 20, the experiment has $C_{20}^5$ equally likely outcomes.

If the batch has 7 defective items, there are 13 good items. The inspector accepts the batch if all 5 selected items are from the 13 good ones. The number of favorable outcomes is $C_{13}^5$.

Therefore $P(A) = \dfrac{C_{13}^5}{C_{20}^5} \approx 8\%$.

Answer: $\dfrac{C_{13}^5}{C_{20}^5} \approx 0.08$.

Statement. In a basketball tournament, 18 teams participate, of which 5 are considered favorites. Teams are divided by drawing into two groups $A$ and $B$ of 9 teams each. What is the probability of: 1) all five favorites landing in the same group (event $M$); 2) exactly two favorites in each group (event $K$)?

Solution. Each group can be formed in $C_{18}^9$ ways.

1. If all 5 favorites go to group $A$, then 4 more teams must be chosen from the remaining 13. This can be done in $C_{13}^4$ ways. Since the 5 favorites could go to either group $A$ or group $B$, the number of favorable outcomes is $2C_{13}^4$. Therefore $P(M) = \dfrac{2C_{13}^4}{C_{18}^9} = \dfrac{1}{34}$.

2. Each group containing exactly 2 favorites can be formed in $C_5^2 \cdot C_{13}^7$ ways. Therefore $P(K) = \dfrac{2 \cdot C_5^2 \cdot C_{13}^7}{C_{18}^9} = \dfrac{12}{17}$.

Answer: 1) $\dfrac{1}{34}$; 2) $\dfrac{12}{17}$.

1. Four letters are chosen at random from the word “ЗАКОН” (Ukrainian for “LAW”). What is the probability that the chosen letters can form the word “КОЗА” (Ukrainian for “GOAT”)?

2. A box contains 10 balls, of which 4 are white. What is the probability that 2 randomly chosen balls are both white?

3. A batch of 100 light bulbs contains 7 defective ones. What is the probability of randomly selecting 4 non-defective bulbs from this batch?