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Sagot :
We have:
- Numbers of balls from 1 to 7 = 7
- Number of balls with number 6 = 1
- Number of ball with number 3 = 1
Then, the probability of ramdomly choosing a 6 is
[tex]p(6)=\frac{1}{7}[/tex]Once we chose a ball, there are 6 balls into the bucket. Then the probability of ramdomly choosing a 3 is
[tex]P(3)=\frac{1}{6}[/tex]Then, the probability of randomly choosing a 6 and 3 in that order, is
[tex]\begin{gathered} P(6\text{and}3)=P(6)\cdot P(3)=\frac{1}{7}\cdot\frac{1}{6} \\ P(6\text{ and 3)=}\frac{1}{7\cdot6} \\ P(6\text{ and 3)=}\frac{1}{42} \end{gathered}[/tex]that is, the probability is 1/42 = 0.023809
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