3 years ago

PLZ HURRY IT'S URGENT!!!!!

Mathematics

1 answer:

Korvikt [17]3 years ago

4 0

Answer:

the answer you are looking for is -11s

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Events a, b, and c in a sample space have p(a)=0.2, p(b)=0.4, and p(c)=0.1. find p(a ∪ b ∪

Julli [10]

If $A$ and $B$ are mutually independent, then $\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)$ and $\mathbb P(A\cap B)=0$ .

If $A$ and $C$ are independent, then $\mathbb P(A\cap C)=\mathbb P(A)\cdot\mathbb P(C)$ . Ditto for $B$ and $C$ .

The inclusion/exclusion principle says that

$\mathbb P(A\cup B\cup C)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\mathbb P(A\cap B)-\mathbb P(A\cap C)-\mathbb P(B\cap C)+\mathbb P(A\cap B\cap C)$

Since $A$ and $B$ are mutually exclusive, their intersection is the empty set. It then follows that $A\cap B\cap C$ also is the empty set, so the last probability is also 0.

Plug in what we know:

$\mathbb P(A\cup B\cup C)=0.2+0.4+0.1-0-0.2\cdot0.1-0.4\cdot0.1+0$

$\implies\mathbb P(A\cup B\cup C)=0.64$

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Nadusha1986 [10]

Answer: i need points ty

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Scorpion4ik [409]

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Juliette [100K]

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Kitty [74]

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Step-by-step explanation:

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3 years ago