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 Concept:
Probability is the measure of the likelihood that an event will occur.
In problems involving "drawing without replacement," the outcome of the first event changes the total number of items and the specific count of items for the second event.
These are known as dependent events.
The total probability of two consecutive dependent events \(A\) and \(B\) is calculated using the formula:
\[ P(A \text{ and } B) = P(A) \times P(B|A) \]
where \(P(B|A)\) is the conditional probability of \(B\) happening given that \(A\) has already occurred.
Step 2: Detailed Explanation:
Let's break down the composition of the bag:
Number of Red balls (\(n_R\)) = \(5\)
Number of Blue balls (\(n_B\)) = \(3\)
Total balls (\(N\)) = \(5 + 3 = 8\)

We need to find the probability that both the first ball drawn and the second ball drawn are red.

Step 2.1: First Draw:
The probability that the first ball is red (\(P(R_1)\)):
Number of favorable outcomes (Red balls) = \(5\)
Total outcomes (Total balls) = \(8\)
\[ P(R_1) = \frac{5}{8} \]

Step 2.2: Second Draw:
Now, one red ball has been removed and not replaced.
Remaining number of red balls = \(5 - 1 = 4\)
Remaining total number of balls = \(8 - 1 = 7\)
The probability that the second ball is red given the first was red (\(P(R_2 | R_1)\)):
\[ P(R_2 | R_1) = \frac{4}{7} \]

Step 2.3: Combined Probability:
\[ P(\text{Both Red}) = P(R_1) \times P(R_2 | R_1) \]
\[ P(\text{Both Red}) = \frac{5}{8} \times \frac{4}{7} \]
\[ P(\text{Both Red}) = \frac{5 \times 4}{8 \times 7} = \frac{20}{56} \]

Step 2.4: Simplification:
Both \(20\) and \(56\) are divisible by \(4\).
\[ \frac{20 \div 4}{56 \div 4} = \frac{5}{14} \]
Step 3: Final Answer:
The probability of drawing two red balls consecutively without replacement is \(\frac{5}{14}\).