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Softa [21]
3 years ago
6

Help.. ~Probability

Mathematics
1 answer:
Alexus [3.1K]3 years ago
5 0

A.

|\Omega|=6\\|A|=3\\\\P(A)=\dfrac{3}{6}=\dfrac{1}{2}

B.

|\Omega|=8\\|A|=3\\\\P(A)=\dfrac{3}{8}

C.

|\Omega|=4\\|A|=1\\\\P(A)=\dfrac{1}{4}

D.

|\Omega|=24\\|A|=5\\\\P(A)=\dfrac{5}{24}

E.

|\Omega|=10\\|A|=10\\\\P(A)=\dfrac{10}{10}=1

F.

|\Omega|=10\\|A|=0\\\\P(A)=\dfrac{0}{10}=0

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OverLord2011 [107]

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