A bag contains 5 red balls and 5 blue balls. If you draw two balls without replacement, what is the probability that the second ball is red, given that the first ball was red?

Question

BlackTom AI · Accepted Answer

Consider the scenario step by step, focusing on how the composition of the bag changes after the first draw.
Option a: 5/10. This would imply no change in the counts after the first red draw, treating the second draw as if the total and red count remained the same. Since we removed one red ball, both the numerator and denominator should reflect the updated totals, so this is incorrect.
Option b: 4/9. After drawing a red ball first, the bag now contains 4 red and 5 blue balls, for a total of 9 balls. The probability that the next ball is red is the number of remaining red balls divided by the new total: 4/9. This matches the updated composition, so this is correct.
Option c: 5/9. This would be the probability if the red count remained at 5 while the total reduced to 9, which cannot happen since one red ball was removed. It miscounts both the remaining red and total, making it incorrect.
Option d: 4/10. This would correspond to 4 red out of 10 total, implying no reduction in the total number of balls, which is not the case after the first draw in a without-replacement scenario. Therefore this is not right.

Statistics and Probability

A bag contains 5 red balls and 5 blue balls. If you draw two balls without replacement, what is the probability that the second ball is red, given that the first ball was red?

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