the track team gives awards for first, second, and third - place runners. there are 10 students from school…

the track team gives awards for first, second, and third - place runners. there are 10 students from school a and 12 students from school b competing. which expression represents the probability that all three awards will go to a student from school b?\n\\(\\frac{_{12}p_{3}}{_{22}p_{3}}\\)\n\\(\\frac{_{12}c_{3}}{_{22}c_{3}}\\)\n\\(\\frac{_{22}p_{3}}{_{22}p_{12}}\\)\n\\(\\frac{_{22}c_{3}}{_{22}c_{12}}\\)

the track team gives awards for first, second, and third - place runners. there are 10 students from school a and 12 students from school b competing. which expression represents the probability that all three awards will go to a student from school b?\n\\(\\frac{_{12}p_{3}}{_{22}p_{3}}\\)\n\\(\\frac{_{12}c_{3}}{_{22}c_{3}}\\)\n\\(\\frac{_{22}p_{3}}{_{22}p_{12}}\\)\n\\(\\frac{_{22}c_{3}}{_{22}c_{12}}\\)

Answer

Answer:

$\frac{{12}P{3}}{{22}P{3}}$

Explanation:

Step1: Calculate total number of students

There are $10$ students from school A and $12$ students from school B, so total students $n = 10 + 12=22$.

Step2: Define probability concept

Probability of an event $E$ is $P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.

Step3: Calculate total number of outcomes

We are awarding first, second and third - place. The total number of ways to choose $3$ students out of $22$ for the three places is given by permutations ${22}P{3}=\frac{22!}{(22 - 3)!}=\frac{22!}{19!}=22\times21\times20$.

Step4: Calculate number of favorable outcomes

The number of ways to choose $3$ students out of $12$ students from school B for the three places is given by permutations ${12}P{3}=\frac{12!}{(12 - 3)!}=\frac{12!}{9!}=12\times11\times10$.

Step5: Find the probability

The probability that all three awards will go to a student from school B is $\frac{{12}P{3}}{{22}P{3}}$.