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What is the probability of rolling a number cube twice and getting an odd number each time?
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1/4
1/2 the rolls are independent
\[\frac{1}{2}\times\frac{1}{2}\]
\[P(\text{odd on }1^{st} \text{ and }\text{odd on }2^{nd})\] \[=P(\text{odd on }1^{st})P(\text{odd on }2^{nd})\] \[=\frac{1}{2}\times\frac{1}{2}\]
1,3,5 are odd numbers that are on the cube
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Thanks Zarkon. You got the answer, it was a simple calculation error. So it is (3/6)*(3/6) = 1/4
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