Question

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

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

Hint:

Using combinations can sometimes save time and prevent sequential multiplication slip-ups: \[ P = \frac{\binom{5}{2}}{\binom{8}{2}} = \frac{\frac{5 \times 4}{2 \times 1}}{\frac{8 \times 7}{2 \times 1}} = \frac{20}{56} = \frac{5}{14} \] Both methods yield identical results, but combinations scale much better if you are asked to select three or more balls at once.

Answer 1

The Correct Option is A

Concept: When drawing objects from a finite set without replacement, the outcomes of the two draws are dependent events. The probability of both events occurring successfully can be evaluated using multiplication rule of probability: \[ P(R_1 \cap R_2) = P(R_1) \times P(R_2 | R_1) \] Alternatively, the problem can be calculated using combinations to choose a subset of objects from the total available group: \[ P = \frac{\text{Number of favorable ways}}{\text{Total number of possible ways}} = \frac{\binom{\text{Red balls}}{2}}{\binom{\text{Total balls}}{2}} \]

Step 1: Determining total balls and analyzing sequential selection.
The bag contains:

• Number of Red balls = 5
• Number of Blue balls = 3
• Total number of balls = 5 + 3 = 8

Let \(R_1\) be the event that the first ball drawn is red, and \(R_2\) be the event that the second ball drawn is red.

Step 2: Calculating the probability step-by-step.
The probability that the first ball is red: \[ P(R_1) = \frac{5}{8} \] Since the ball is drawn without replacement, the number of remaining red balls becomes \(5 - 1 = 4\), and the total number of remaining balls in the bag drops to \(8 - 1 = 7\). The conditional probability that the second ball is red, given the first was red: \[ P(R_2 | R_1) = \frac{4}{7} \] Applying the multiplication rule for dependent events: \[ P(\text{Both are red}) = P(R_1) \times P(R_2 | R_1) = \frac{5}{8} \times \frac{4}{7} \] Simplifying the fractions by cancelling out common factors: \[ P(\text{Both are red}) = \frac{5 \times 1}{2 \times 7} = \frac{5}{14} \]