Question

There are 3 chess players in the 1st year, 4 chess players in the 2nd year, and 5 chess players in the 3rd year of college. What is the probability of selecting a team of 4 chess players if at least one player is from the 1st year?

• A. 41/55
• B. 27/110
• C. 19/66
• D. 37/60
• E. 82/99

Hint:

In probability problems, use the complement rule when the condition involves at least one of a certain type.

Answer 1

The Correct Option is A

Step 1: Calculate the total number of possible teams.
There are \( 3 + 4 + 5 = 12 \) chess players in total. The number of ways to choose 4 players from 12 is: \[ \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495. \] Step 2: Calculate the number of teams with no 1st-year players.
The number of ways to select 4 players from the 2nd and 3rd years (a total of \( 4 + 5 = 9 \) players) is: \[ \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126. \] Step 3: Calculate the number of successful teams (at least one 1st-year player).
Subtract the number of teams with no 1st-year players from the total number of teams: \[ 495 - 126 = 369. \] Step 4: Determine the probability.
The probability of selecting a team with at least one 1st-year player is: \[ \frac{369}{495}. \] Simplifying the fraction: \[ \frac{369}{495} = \frac{41}{55}. \] Step 5: Final Answer.
The probability is \( \frac{41}{55} \).