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IrinaK [193]
2 years ago
14

Given P(A) = 0.072, P(B) = 0.180, and P(C) = 0.027, and that events A, B, and C are mutually exclusive, what is the P(A or B or

C)?
Mathematics
1 answer:
9966 [12]2 years ago
5 0

Given P(A) = 0.072, P(B) = 0.180, and P(C) = 0.027, and that events A, B, and C are mutually exclusive,then P(A or B or C) is 0.279

<u>Solution: </u>

Given that probability of event A is P(A) = 0.072

Probability of event B is P(B) = 0.180

Probability of event C is P(C) = 0.027

Also event A, B and C are mutually exclusive.

Need to determine P( A or B or C )

For mutually exclusive events  

P( A or B or C ) = P(A) + P(B) + P(C)

= 0.072 + 0.180 + 0.027  = 0.279

Hence probability of occurrence of event A or event B or event C , where A , B and C are mutually exclusive events is 0.279.

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2 years ago
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Answer:

If the time passed is only 3 months, then it is $2040

Step-by-step explanation:

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Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the even
Allushta [10]

Answer:

P(E|A)= \frac{10}{11}

Step-by-step explanation:

Given

Two rolls of die

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Required

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First, list out the outcome of each

E = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}

A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}

So:

P(E|A)= \frac{n(E\ n\ A)}{n(A)}

Where:

E\ n\ A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}

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