Question

A bag contains 5 red balls and 3 green balls. If two balls are drawn at random without replacement, what is the probability that both balls drawn are red?

• A. \( \frac{5}{28} \)
• B. \( \frac{5}{21} \)
• C. \( \frac{3}{14} \)
• D. \( \frac{1}{3} \)

Hint:

When calculating probabilities in dependent events (like drawing without replacement), multiply the probabilities of each step. Always adjust the total number of outcomes after each draw.

Answer 1

The Correct Option is B

We are given a bag containing 5 red balls and 3 green balls, and we are asked to find the probability of drawing two red balls without replacement. Step 1: Total number of balls The total number of balls in the bag is: \[ 5 \, \text{(red balls)} + 3 \, \text{(green balls)} = 8 \, \text{balls} \] Step 2: Probability of drawing the first red ball The probability of drawing the first red ball is: \[ P(\text{First red}) = \frac{5}{8} \] Step 3: Probability of drawing the second red ball After drawing the first red ball, there are 4 red balls left and the total number of balls is now 7. Thus, the probability of drawing the second red ball is: \[ P(\text{Second red}) = \frac{4}{7} \] Step 4: Multiply the probabilities Since the events are dependent (we are drawing without replacement), the probability that both balls drawn are red is the product of the individual probabilities: \[ P(\text{Both red}) = P(\text{First red}) \times P(\text{Second red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{21} \] Answer: The probability that both balls drawn are red is \( \frac{5}{21} \), so the correct answer is option (2).