In a prize drawing, the tokens that participants will randomly select are numbered from to . of

Question

these tokens are blue, are yellow, and are green. The first prize goes to the participant who
draws of the same color in a row without replacing tokens after each draw. What is the
probability of drawing of the same color?

Answer 1

To determine the probability of drawing \( k \) tokens of the same color in a row, we need to consider a few key pieces of information:

1. The total number of tokens.
2. The distribution of colors among the tokens.
3. The specific value of \( k \) (the number of consecutive tokens drawn of the same color).

Let's denote:

• The total number of tokens as \( N \).
• The number of blue tokens as \( B \).
• The number of yellow tokens as \( Y \).
• The number of green tokens as \( G \).

Now we want to calculate the probability of drawing \( k \) tokens of the same color in sequence.

The probability for drawing \( k \) tokens of the same color can be calculated as follows:

1. Probability of drawing \( k \) blue tokens: \[ P(B, k) = \frac{B}{N} \times \frac{B-1}{N-1} \times \frac{B-2}{N-2} \times \ldots \times \frac{B-k+1}{N-k+1} \]

2. Probability of drawing \( k \) yellow tokens: \[ P(Y, k) = \frac{Y}{N} \times \frac{Y-1}{N-1} \times \frac{Y-2}{N-2} \times \ldots \times \frac{Y-k+1}{N-k+1} \]

3. Probability of drawing \( k \) green tokens: \[ P(G, k) = \frac{G}{N} \times \frac{G-1}{N-1} \times \frac{G-2}{N-2} \times \ldots \times \frac{G-k+1}{N-k+1} \]

4. Total probability of drawing \( k \) tokens of the same color: \[ P(k) = P(B, k) + P(Y, k) + P(G, k) \]

In summary, the overall probability involves calculating the individual probabilities for each color, summing them, and requires knowing the total count of each color to substitute in the formulas above.

If you provide the total number of tokens \( N \), along with the specific counts of blue, yellow, and green tokens, I can perform the calculations for you. Also, please specify the value of \( k \).