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: Total number of ways to select 4 players from all the players.
There are a total of \( 3 + 4 + 5 = 12 \) chess players. The total number of ways to select 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: Number of ways to select 4 players with no 1st-year player.
The number of ways to select 4 players from the 2nd and 3rd years (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: Number of favorable outcomes.
The number of favorable outcomes where at least one player is from the 1st year is the total number of ways minus the number of ways with no 1st-year player: \[ 495 - 126 = 369. \] Step 4: Probability.
The probability of selecting a team of 4 players with at least one 1st-year player is: \[ \frac{369}{495}. \] Simplifying this fraction: \[ \frac{369}{495} = \frac{41}{55}. \] Step 5: Conclusion.
The probability is \( \frac{41}{55} \).