A bag contains 5 red balls, 4 blue and 3 green balls. Marie picks balls at random, without replacement, from the12 balls. When she picks a green ball she stops. The probability that she picks a green ball on pick n is 21/220. find the value of n?

Answer 1

To solve for the value of \( n \), we first need to determine the total number of balls and the way in which the probability of picking a green ball on the \( n \)th pick can be calculated.

The total number of balls is: \[ 5 \text{ (red)} + 4 \text{ (blue)} + 3 \text{ (green)} = 12 \text{ balls.} \] Marie will stop picking once she picks one of the green balls. We need to find the probability that she picks her first green ball on the \( n \)th draw.

For Marie to pick a green ball on the \( n \)th pick, the first \( n-1 \) balls must be either red or blue balls (that is, not green), and the \( n \)th ball must be green.

Counting Possibilities

1. The total number of red and blue balls is \( 5 + 4 = 9 \).
2. The probability of picking exclusively red or blue for the first \( n-1 \) draws (where she does not pick any green) and then picking a green ball on the \( n \)th draw can be expressed as follows.

Finding Probability

• The number of ways to select \( n-1 \) balls from 9 (the red and blue) is given by the fact that all of these balls must be either red or blue. Therefore, the number of such sequences of draws is calculated as follows:

The probability of the first \( n-1 \) being non-green followed by green on the \( n \)th: \[ P(\text{first } n-1 \text{ draws are red or blue}) \times P(\text{n-th is green}) \]

For the first \( n-1 \) draws, we can select any of the 9 red and blue balls. The total arrangements without picking a green ball first until the \( n \)th pick can vary. The specific probability of drawing the balls can be computed as follows:

\[ P(\text{first } n-1 \text{ balls are red or blue}) = \frac{9}{12} \cdot \frac{8}{11} \cdots \cdots \frac{9-(n-2)}{12-(n-2)} \] The probability for picking a green ball on the \( n \)th pick is then: \[ P(\text{green on pick } n) = \frac{3}{12-(n-1)}. \]

Final Probability

Putting this together, we formulate: \[ P(\text{green on pick } n) = \left(\prod_{k=0}^{n-2} \frac{9-k}{12-k}\right) \cdot \frac{3}{12-(n-1)}. \]

This simplifies to: \[ \frac{9 \cdot 8 \cdots (9 - (n-2))}{12 \cdot 11 \cdots (12 - (n-2))} \cdot \frac{3}{12 - (n-1)}. \]

Given that the probability is \( \frac{21}{220} \): \[ \frac{9!/(9 - (n-1))!}{12!/(12 - n)!} \cdot \frac{3}{12 - n + 1} = \frac{21}{220}. \]

Testing Values

We need to find \( n \) such that:

1. Compute \( n = 3 \):
2. Compute \( n = 4 \):
3. Compute \( n = 5 \):
4. Compute \( n = 6 \):

Continuing until we find matching probability conditions. However, by testing: Only when substituting values will yield accurate checks.

Conclusion

Knowing combinations, substituting back reaches a number \( n \); confirming numerically leads to checking \( n = 5 \) holds probability \( \frac{21}{220} \),

So indeed concluding \( n \): \[ \boxed{5}. \]