Since ∣S∣=10|S|=10∣S∣=10 and the outcomes are equally likely, then
P({i})=110P(\{i\})=\frac{1}{10}P({i})=101 for i=1,...,10i=1,...,10i=1,...,10.
Since ∣E∣=3|E|=3∣E∣=3, then P(E)=3⋅110=0.3.P(E)=3\cdot\frac{1}{10}=0.3.P(E)=3⋅101=0.3.
Answer: P(E)=0.3P(E)=0.3P(E)=0.3.
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