Answer :
The probability of picking a black checker, replacing it, and then picking another black checker from a bag of six black checkers and four red checkers is (6/10) * (6/10) = 36/100 = 0.36 or 36%.
In the first pick, there are six black checkers out of a total of ten checkers (six black checkers and four red checkers), so the probability of picking a black checker is 6/10 or 0.6.
Since the checker is replaced after the first pick, the bag remains the same for the second pick. In the second pick, the probability of picking another black checker is also 6/10 or 0.6.
To find the overall probability, we multiply the probabilities of the individual events, resulting in 0.6 * 0.6 = 0.36 or 36%.
To explain further, the probability is determined by dividing the number of favorable outcomes by the total number of possible outcomes. In this case, the favorable outcome is picking a black checker twice, and there are six black checkers available for each pick.
The total number of possible outcomes is the total number of checkers in the bag, which is ten. By multiplying the probabilities of the two independent events (picking a black checker on the first and second picks), we obtain the probability of both events occurring consecutively. Thus, the probability of picking a black checker, replacing it, and then picking another black checker is 0.36 or 36%.
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