Question

write and solve an original problem involving probability and combinations whose solution requires $\frac{_{12}c_{9}}{_{21}c_{9}}$. write a problem involving probability and combinations whose solution requires $\frac{_{12}c_{9}}{_{21}c_{9}}$. a. a musical talent show consists of 12 string orchestras and 9 wind orchestras. what is the probability that 9 string orchestras will perform in a row? b. a bag contains 12 black marbles and 9 white marbles. what is the probability that 9 black marbles are chosen at random? c. a hockey tryout consists of 12 left - handed shooters and 9 right - handed shooters. if 9 players will make the team, what is the probability that they are all left - handed shooters? d. a tree farm consists of 12 douglas fir and 9 spruce trees. if a hotel wants to purchase 9 random trees, what is the probability that they are all spruce trees? solve the problem whose solution requires $\frac{_{12}c_{9}}{_{21}c_{9}}$. the probability is (type an integer or a simplified fraction.)

Explanation:

Step1: Recall combination formula

The combination formula is $_{n}C_{r}=\frac{n!}{r!(n - r)!}$. Here, the probability of an event $E$ is given by $P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. The total number of items is $12 + 9=21$, and we are choosing 9 items. The number of favorable - outcomes is the number of ways to choose 9 items from 12, and the total number of outcomes is the number of ways to choose 9 items from 21.

Step2: Analyze each option

• Option A: The problem is about the order of performance, and combination formula is for non - ordered selection, so this is incorrect.
• Option B: The total number of marbles is $12 + 9 = 21$, and we want to choose 9 black marbles from 12 black marbles. The number of favorable outcomes is $_{12}C_{9}$, and the total number of outcomes is $_{21}C_{9}$. This option is correct.
• Option C: The total number of players is $12+9 = 21$, but we want all left - handed shooters. The number of favorable outcomes should be choosing 9 from 12 left - handed shooters, and the total number of outcomes is choosing 9 from 21 players. But the description in the option is about making the team, which is not in line with the pure combination - based probability concept as required by the formula in the most straightforward way.
• Option D: We want all spruce trees, but we should be choosing from 9 spruce trees, and the formula $_{12}C_{9}$ in the numerator does not match this situation.

Step3: Calculate the combination values

$_{n}C_{r}=\frac{n!}{r!(n - r)!}$, so $_{12}C_{9}=_{12}C_{3}=\frac{12!}{3!(12 - 3)!}=\frac{12\times11\times10}{3\times2\times1}=220$ and $_{21}C_{9}=\frac{21!}{9!(21 - 9)!}=\frac{21!}{9!12!}=\frac{21\times20\times19\times18\times17\times16\times15\times14\times13}{9\times8\times7\times6\times5\times4\times3\times2\times1}=293930$. Then $\frac{_{12}C_{9}}{_{21}C_{9}}=\frac{220}{293930}=\frac{22}{29393}$.

Answer:

B. A bag contains 12 black marbles and 9 white marbles. What is the probability that 9 black marbles are chosen at random?; $\frac{22}{29393}$