Probability of Drawing at Least One Red Ball and At Most One Green Ball from a Bag

Objective: Calculate the probability of drawing at least one red ball and at most one green ball from a bag containing 4 red, 3 black, and 3 green balls.

Step-by-Step Solution

Total Number of Balls

Total Outcomes

Calculate the total number of ways to choose 2 balls from 10 using the combination formula (C(n, k) frac{n!}{k!(n-k)!}).

Calculation: (C_{10}^{2} frac{10 times 9}{2 times 1} 45)

Favorable Outcomes

To find the probability of drawing at least one red ball and at most one green ball, we can break down the cases as follows:

Case 1: 2 Red Balls

(C_{4}^{2} frac{4 times 3}{2 times 1} 6)

Case 2: 1 Red and 1 Black

(C_{4}^{1} times C_{3}^{1} 4 times 3 12)

Case 3: 1 Red and 1 Green

(C_{4}^{1} times C_{3}^{1} 4 times 3 12)

Total Favorable Outcomes

Sum of favorable outcomes from the cases: (6 12 12 30)

Probability Calculation

The probability P is given by:

(P frac{text{Number of favorable outcomes}}{text{Total outcomes}} frac{30}{45} frac{2}{3})

Final Answer

The probability of drawing at least one red ball and at most one green ball when two balls are drawn is (frac{2}{3}).

Alternative Approach

Alternatively, if two balls are drawn and at least one red ball is drawn, then automatically at most one green ball is drawn. This means:

1 - (frac{6}{10} times frac{5}{9} frac{2}{3})

Conclusion

This approach simplifies the problem by directly calculating the probability of the complementary event and subtracting it from 1.

Keywords: probability, drawing balls, combinatorics