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In two independent throws of a die, how do you know how many ways that you are going to have or not for certain number ?? For example, if the number was 4, there are number of ways to have none of 4's, there are number of ways to have just one 4's, and there are number of ways to have two 4's. so (1,1) would be one of ways to have none of 4's |
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I read your question as follows: what are, when throwing a fair dice twice, the probabilities... P(2 4s) = $\frac{1}{6} \frac{1}{6} = \frac{1}{36}$ P(1 4) = P(4 in first but not second throw) + P(4 in second but not first throw) = (by symmetry) $2 \frac{1}{6} \frac{5}{6} = \frac{10}{36}$ P(no 4) = $\frac{5}{6} \frac{5}{6} = \frac{25}{36}$, noting that the 3 probabilities sum to $1$, as they should. $4$ is not special, so same result for any of the $6$ numbers. |
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Draw a table with 6 rows and 6 columns. Number rows and columns from 1 to 6. In the cells write the sum of row and column number. Look at the occurences of the number 4: where they dispose? How many 4 are there? |
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