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Sagot :
In order to have independent events, we need to verify the following formula:
[tex]P(A\cap B)=P(A)\cdot P(B)[/tex]For event A, the outcomes that are less than 6 are 1, 2, 3, 4 and 5, so the probability of event A is:
[tex]\begin{gathered} P(A)=P(1)+P(2)+P(3)+P(4)+P(5) \\ P(A)=0.1+0.2+0.2+0.2+0.2 \\ P(A)=0.9 \end{gathered}[/tex]For event B, the outcomes that are divisible by 3 are 3 and 6, so:
[tex]\begin{gathered} P(B)=P(3)+P(6) \\ P(B)=0.2+0.1 \\ P(B)=0.3 \end{gathered}[/tex]The intersection of these events is only number 3, so:
[tex]P(A\cap B)=P(3)=0.2[/tex]Verifying the formula, we have:
[tex]\begin{gathered} 0.2=0.9\cdot0.3 \\ 0.2=0.27 \end{gathered}[/tex]The result is false, so the events are not independent, therefore the answer is NO.
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