Question

A bag contains 5 Red and 6 White balls. A person draws two balls one after the other without replacing the first one. The probability that both balls will be of same color will be

• A. 2/11
• B. 25/30
• C. 5/11
• D. 25/36

Hint:

The keyword "without replacing" is crucial. It signals that the total number of outcomes for the second event changes. If it were "with replacement," the denominator would remain 11 for the second draw.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a problem of dependent probability, as the outcome of the second draw depends on the first draw (since the ball is not replaced). We need to find the probability of two mutually exclusive events: either drawing two red balls OR drawing two white balls, and then sum their probabilities.

Step 2: Key Formula or Approach:
Total number of balls = 5 Red + 6 White = 11 balls.
The probability of both balls being the same color is P(Both Red) + P(Both White).
For dependent events, P(A and B) = P(A) * P(B|A), where P(B|A) is the probability of B occurring given that A has already occurred.

Step 3: Detailed Explanation:
Case 1: Probability of drawing two Red balls (RR)
Probability of drawing the first red ball: \( P(R_1) = \frac{\text{Number of Red balls}}{\text{Total balls}} = \frac{5}{11} \)
After drawing one red ball, there are 4 red balls left and a total of 10 balls.
Probability of drawing the second red ball: \( P(R_2|R_1) = \frac{\text{Remaining Red balls}}{\text{Remaining Total balls}} = \frac{4}{10} \)
Probability of both being red: \( P(RR) = P(R_1) \times P(R_2|R_1) = \frac{5}{11} \times \frac{4}{10} = \frac{20}{110} \)
Case 2: Probability of drawing two White balls (WW)
Probability of drawing the first white ball: \( P(W_1) = \frac{\text{Number of White balls}}{\text{Total balls}} = \frac{6}{11} \)
After drawing one white ball, there are 5 white balls left and a total of 10 balls.
Probability of drawing the second white ball: \( P(W_2|W_1) = \frac{\text{Remaining White balls}}{\text{Remaining Total balls}} = \frac{5}{10} \)
Probability of both being white: \( P(WW) = P(W_1) \times P(W_2|W_1) = \frac{6}{11} \times \frac{5}{10} = \frac{30}{110} \)
Total Probability:
The total probability of drawing two balls of the same color is the sum of the probabilities of these two mutually exclusive cases.
\[ P(\text{Same Color}) = P(RR) + P(WW) = \frac{20}{110} + \frac{30}{110} = \frac{50}{110} \] Simplifying the fraction:
\[ \frac{50}{110} = \frac{5}{11} \]

Step 4: Final Answer:
The probability that both balls will be of the same color is 5/11.