A bag contains 5 red, 7 green, and 3 blue balls. Two balls are drawn at random from the bag one-by-one. The probability of the second drawn ball being red is:
Show Hint
- \( \frac{1}{5} \)
- \( \frac{1}{3} \)
- \( \frac{2}{5} \)
- \( \frac{2}{3} \)
The Correct Option is B
Solution and Explanation
We are given a bag containing 5 red, 7 green, and 3 blue balls. The total number of balls in the bag is:
\[ 5 + 7 + 3 = 15 \]
We are tasked with finding the probability that the second drawn ball is red. The key here is to consider the possible cases that can occur when drawing two balls sequentially.
There are two primary cases to analyze: the case where the first ball drawn is red and the case where the first ball drawn is not red.
Case 1: First ball drawn is red.
If the first ball drawn is red, there are now 4 red balls remaining out of the 14 balls left. The probability of drawing a red ball in this case is:
\[ P({second red | first red}) = \frac{4}{14} = \frac{2}{7} \] Case 2: First ball drawn is not red.
If the first ball drawn is not red (i.e., it is either green or blue), there are still 5 red balls remaining out of the 14 balls left. The probability of drawing a red ball in this case is:
\[ P({second red | first not red}) = \frac{5}{14} \] Total probability:
The total probability is found by multiplying the probability of each case by the probability of the first ball being red or not red. The probability of drawing a red ball first is \( \frac{5}{15} \), and the probability of drawing a non-red ball first is \( \frac{10}{15} \). Therefore, the total probability of the second drawn ball being red is:
\[ P({second red}) = \left( \frac{5}{15} \times \frac{2}{7} \right) + \left( \frac{10}{15} \times \frac{5}{14} \right) = \frac{1}{3} \] Thus, the probability that the second ball drawn is red is \( \frac{1}{3} \), corresponding to option (B).
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