Question

A bag contains \(5\) red balls and \(3\) blue balls. Two balls are drawn at random without replacement. What is the probability that both balls are red?

• A. \(\dfrac{5}{28}\)
• B. \(\dfrac{10}{28}\)
• C. \(\dfrac{15}{56}\)
• D. \(\dfrac{5}{14}\)

Hint:

For draws without replacement: \[ P(A\cap B)=P(A)\times P(B|A) \] Always reduce the total number of objects after each draw.

Answer 1

The Correct Option is A

Concept: When objects are drawn without replacement, probabilities change after each draw because the total number of objects decreases. Probability of an event is: \[ P(E)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}} \] For successive events without replacement, probabilities are multiplied.

Step 1: Find the total number of balls. The bag contains: \[ 5 \text{ red balls} \] and \[ 3 \text{ blue balls} \] Therefore total balls: \[ 5+3=8 \]

Step 2: Find probability that the first ball is red. Number of red balls initially: \[ 5 \] Total balls initially: \[ 8 \] Thus, \[ P(\text{first red}) = \frac58 \]

Step 3: Find probability that the second ball is also red. Since one red ball has already been removed: \[ \text{Remaining red balls}=4 \] and \[ \text{Remaining total balls}=7 \] Hence, \[ P(\text{second red} \mid \text{first red}) = \frac47 \]

Step 4: Multiply the probabilities. \[ P(\text{both red}) = \frac58\times\frac47 \] \[ = \frac{20}{56} \] \[ = \frac{5}{14} \] Thus, \[ \boxed{\frac{5}{14}} \] Hence the correct answer is: \[ \boxed{(D)\ \frac{5}{14}} \] Note: The provided answer key appears incorrect. The mathematically correct probability is: \[ \boxed{\frac{5}{14}} \]