Question

A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. The probability that both are red is:

• A. (\(\frac{5}{14}\))
• B. (\(\frac{8}{56}\))
• C. (\(\frac{5}{28}\))
• D. (\(\frac{15}{56}\))

Hint:

For probability problems involving "without replacement," remember that the denominator (total items) and sometimes the numerator (desired items) decrease after each selection. Combinations provide a reliable way to solve such problems. Simplify fractions to their lowest terms for comparison with options.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the probability of drawing two red balls consecutively from a bag without replacement.

Step 2: Key Formula or Approach:
The probability of multiple events occurring in sequence without replacement is calculated by multiplying the probabilities of each individual event, where the sample space decreases after each draw.
Alternatively, using combinations: The probability of selecting a specific combination of items is $\frac{\text{Number of ways to choose desired items}}{\text{Total number of ways to choose items}}$.

Step 3: Detailed Explanation:
Given:
- Number of red balls = 5.
- Number of blue balls = 3.
- Total number of balls = $5 + 3 = 8$.
Method 1: Sequential Probability
1. Probability of drawing the first red ball:
There are 5 red balls out of 8 total.
$P(\text{1st red}) = \frac{5}{8}$.
2. Probability of drawing the second red ball (without replacement):
After drawing one red ball, there are now 4 red balls left and a total of 7 balls.
$P(\text{2nd red | 1st red}) = \frac{4}{7}$.
3. Probability that both are red:
$P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red | 1st red})$
$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$.
Simplify the fraction: $\frac{20}{56} = \frac{5 \times 4}{14 \times 4} = \frac{5}{14}$.
Method 2: Using Combinations
1. Total number of ways to draw 2 balls from 8:
\[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]
2. Number of ways to draw 2 red balls from 5 red balls:
\[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]
3. Probability that both are red:
$P(\text{both red}) = \frac{\text{Number of ways to draw 2 red balls}}{\text{Total number of ways to draw 2 balls}} = \frac{10}{28}$.
Simplify the fraction: $\frac{10}{28} = \frac{5 \times 2}{14 \times 2} = \frac{5}{14}$.

Step 4: Final Answer:
The probability that both balls drawn are red is $\frac{5}{14}$. This corresponds to option (A).