Question:
A bag contains 10 red balls and 15 blue balls. Two balls are drawn randomly without replacement. Given that the first ball drawn is red, the probability (rounded off to 3 decimal places) that both balls drawn are red is \_\_\_\_.
A bag contains 10 red balls and 15 blue balls. Two balls are drawn randomly without replacement. Given that the first ball drawn is red, the probability (rounded off to 3 decimal places) that both balls drawn are red is \_\_\_\_.
Updated On: Jan 22, 2025
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Solution and Explanation
The problem involves calculating the probability \( P(R_2 \mid R_1) \), where \( R_1 \) and \( R_2 \) represent the drawing of red balls in the first and second draws, respectively.
Step 1: Analyze the outcomes
Total number of balls: \( 25 \) (10 Red + 15 Blue).
Probability of drawing \( R_1 \) (a red ball in the first draw): \[ P(R_1) = \frac{10}{25}. \] Step 2: Conditional probabilities
If \( R_1 \) is drawn, there are now \( 9 \) red balls and \( 15 \) blue balls left in the bag (total: \( 24 \)).
The probability of drawing \( R_2 \) (a red ball in the second draw given \( R_1 \)): \[ P(R_2 \mid R_1) = \frac{9}{24}. \] Step 3: Final calculation \[ P(R_2 \mid R_1) = 0.375. \] \section*{Final Answer} \[ \boxed{0.375} \]
Total number of balls: \( 25 \) (10 Red + 15 Blue).
Probability of drawing \( R_1 \) (a red ball in the first draw): \[ P(R_1) = \frac{10}{25}. \] Step 2: Conditional probabilities
If \( R_1 \) is drawn, there are now \( 9 \) red balls and \( 15 \) blue balls left in the bag (total: \( 24 \)).
The probability of drawing \( R_2 \) (a red ball in the second draw given \( R_1 \)): \[ P(R_2 \mid R_1) = \frac{9}{24}. \] Step 3: Final calculation \[ P(R_2 \mid R_1) = 0.375. \] \section*{Final Answer} \[ \boxed{0.375} \]
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