A box contains identical balls of which 2 are red, 16 are white and 8 are blue. 3 balls are drawn from the box one after the other without replacement. Find the probability in which A. 3 are red B. The first is blue and the other 2 are red C. Two are white and 1 is blue

We can approach this problem using the concept of probability and combinations since the balls are drawn without replacement.

### Given:
- 2 red balls
- 16 white balls
- 8 blue balls
- Total balls = \( 2 + 16 + 8 = 26 \)

### Part A: Probability that all 3 balls are red

Since there are only 2 red balls, it is **impossible** to draw 3 red balls. So, the probability is:

\[
P(\text{3 red}) = 0
\]

### Part B: Probability that the first ball is blue and the other 2 are red

- There are 8 blue balls out of 26 total balls, so the probability of drawing the first ball blue is:

\[
P(\text{first is blue}) = \frac{8}{26}
\]

- After drawing a blue ball, 25 balls remain, and only 2 of them are red. The probability of drawing a red ball next is:

\[
P(\text{second is red}) = \frac{2}{25}
\]

- After drawing one red ball, 24 balls remain, and only 1 of them is red. The probability of drawing the second red ball is:

\[
P(\text{third is red}) = \frac{1}{24}
\]

So, the total probability is the product of all these individual probabilities:

\[
P(\text{blue, red, red}) = \frac{8}{26} \times \frac{2}{25} \times \frac{1}{24}
\]

Let’s calculate that:

\[
P(\text{blue, red, red}) = \frac{8 \times 2 \times 1}{26 \times 25 \times 24} = \frac{16}{15600} = \frac{2}{1950}
\]

### Part C: Probability that 2 balls are white and 1 ball is blue

- There are 16 white balls out of 26, so the probability of drawing the first white ball is:

\[
P(\text{first white}) = \frac{16}{26}
\]

- After drawing a white ball, 25 balls remain, and 15 of them are white. The probability of drawing the second white ball is:

\[
P(\text{second white}) = \frac{15}{25}
\]

- After drawing 2 white balls, 24 balls remain, and 8 of them are blue. The probability of drawing the blue ball next is:

\[
P(\text{third blue}) = \frac{8}{24}
\]

So, the total probability is the product of these individual probabilities:

\[
P(\text{2 white, 1 blue}) = \frac{16}{26} \times \frac{15}{25} \times \frac{8}{24}
\]

Let’s calculate that:

\[
P(\text{2 white, 1 blue}) = \frac{16 \times 15 \times 8}{26 \times 25 \times 24} = \frac{1920}{15600} = \frac{4}{65}
\]

### Final Answers:
A. Probability of drawing 3 red balls: \( 0 \)

B. Probability of drawing the first ball blue and the other two red: \( \frac{2}{1950} \)

C. Probability of drawing 2 white balls and 1 blue ball: \( \frac{4}{65} \)

---

Do you have any questions or would you like further explanation?

Here are 5 related questions to expand your understanding:

1. What is the probability of drawing 3 blue balls?
2. How would the probability change if there were 10 blue balls instead of 8?
3. What is the probability of drawing 2 red balls and 1 white ball?
4. How would the results differ if the balls were drawn with replacement?
5. What is the probability that none of the 3 drawn balls is red?

**Tip:** When calculating probabilities without replacement, always remember to adjust the total number of available items after each draw!

Learn how to calculate probabilities in a scenario where a box contains 2 red, 16 white, and 8 blue balls, and 3 balls are drawn without replacement. This problem involves key concepts in probability and combinatorics, with solutions for different cases such as drawing 3 red balls or 2 white and 1 blue. Suitable for students in Grades 9-12.

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