Question

g. find the probability of drawing a blue block, keeping it, and then drawing a red block.

Explanation:

Step1: Define Variables (Assume Totals)

Let’s assume there are \( B \) blue blocks, \( R \) red blocks, and total blocks \( T = B + R + \text{other colors} \). But since we don't have numbers, we'll use general probability rules. The first event: probability of blue is \( P(\text{blue})=\frac{B}{T} \).

Step2: Update Total After First Draw

After keeping the blue block, total blocks become \( T - 1 \), and red blocks remain \( R \). So probability of red next is \( P(\text{red}|\text{blue})=\frac{R}{T - 1} \).

Step3: Multiply Probabilities (Dependent Events)

Since the events are dependent (we keep the first block), the combined probability is \( P = \frac{B}{T} \times \frac{R}{T - 1} \). For example, if there are 3 blue, 5 red, and 10 total: \( P=\frac{3}{10}\times\frac{5}{9}=\frac{15}{90}=\frac{1}{6} \). But without specific counts, the formula is \( \frac{\text{Number of blue}}{\text{Total}} \times \frac{\text{Number of red}}{\text{Total}-1} \).

Answer:

To find the probability, follow these steps (using \( B \) = blue blocks, \( R \) = red blocks, \( T \) = total blocks):

1. Probability of blue first: \( \boldsymbol{\frac{B}{T}} \) (since we keep it, total decreases by 1).
2. Probability of red next: \( \boldsymbol{\frac{R}{T - 1}} \) (total is now \( T - 1 \), red count remains \( R \)).
3. Multiply the two probabilities: \( \boldsymbol{\frac{B}{T} \times \frac{R}{T - 1}} \).

For example, if \( B = 3 \), \( R = 5 \), \( T = 10 \): \( \frac{3}{10} \times \frac{5}{9} = \frac{1}{6} \). The final answer depends on the specific number of blue, red, and total blocks.