There are a total of "3+4+5=12"
The probability of picking one red ball is "\\frac{3}{12}= \\frac{1}{4}" . Since the first ball is replaced, the probability of picking a red ball during the second picking is "\\frac{1}{4}"
Thus, the probability of picking two red balls with replacement is "\\frac{1}{4}\u00d7\\frac{1}{4}= \\frac{1}{16}"
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