[tex]\displaystyle|\Omega|=2^4=16\\|A|=\binom{4}{3}=\dfrac{4!}{3!1!}=4\\\\P(A)=\dfrac{4}{16}=\dfrac{1}{4}[/tex]
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[tex]\displaystyle|\Omega|=2^4=16\\|A|=\binom{4}{3}=\dfrac{4!}{3!1!}=4\\\\P(A)=\dfrac{4}{16}=\dfrac{1}{4}[/tex]