To determine the probability that all three awards will go to students from school B, we need to consider the probabilities of selecting students from school B in three successive selections without replacement.
Here are the steps for calculating this probability:
1. Total Students: The total number of competing students is 10 from school A plus 12 from school B, which sums up to 22 students in total.
2. First Selection: The probability that the first student selected is from school B is [tex]\( \frac{12}{22} \)[/tex].
3. Second Selection: After one student from school B has been chosen, 11 students from school B remain out of the remaining 21 students. So, the probability that the second student selected is also from school B is [tex]\( \frac{11}{21} \)[/tex].
4. Third Selection: After two students from school B have been chosen, 10 students from school B remain out of the remaining 20 students. Therefore, the probability that the third student selected is from school B is [tex]\( \frac{10}{20} \)[/tex] or [tex]\( \frac{1}{2} \)[/tex].
To find the overall probability that all three selections are from school B, we multiply these individual probabilities together:
[tex]\[ \left( \frac{12}{22} \right) \times \left( \frac{11}{21} \right) \times \left( \frac{10}{20} \right) \][/tex]
Now, let's analyze the given options:
1. [tex]\(\frac{12^P}{22^P}\)[/tex] - This option is not correct as it does not represent the probability of selecting students from school B specifically.
2. [tex]\(\frac{{ }_{12} C_3}{{ }_{22} C_3}\)[/tex] - This represents the ratio of the number of ways to choose 3 students out of 12 to the number of ways to choose 3 students out of 22. While related to combinations, this is not the specific sequential probability calculation.
3. [tex]\(\frac{22 P_3}{22 P_{12}}\)[/tex] - This expression is also not correct for the scenario of choosing three students consecutively.
4. [tex]\(\frac{{ }_{22} C _3}{{ }_{22} C _{12}}\)[/tex] - This is an incorrect interpretation for the required probability.
The correct answer is the approach detailed in the steps above, involving the probabilities of successive selections:
[tex]\[ \left( \frac{12}{22} \right) \times \left( \frac{11}{21} \right) \times \left( \frac{10}{20} \right) \][/tex]
This multiplication gives us the probability [tex]\(0.14285714285714285\)[/tex], which matches our objective calculation.
