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Re: The probability that Maria will eat breakfast on any given d [#permalink]
i'm confused here where is the double counts happen? the both events are independent.
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Re: The probability that Maria will eat breakfast on any given d [#permalink]
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GR3A wrote:
i'm confused here where is the double counts happen? the both events are independent.


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Plz see the diag above.

The intersection i.e probability of Event A & Event B is counted twice.

So in OR probability of event A or event B :- P(A) + P(B) - P(A & B)
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Re: The probability that Maria will eat breakfast on any given d [#permalink]
GR3A wrote:
i'm confused here where is the double counts happen? the both events are independent.


I believe you may be confusing independence with mutual exclusivity.
If two events are mutually exclusive then they cannot both occur at the same time.

Cheers,
Brent
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Re: The probability that Maria will eat breakfast on any given d [#permalink]
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We can solve this problem using M1 (probability that Maria eats breakfast) = 0.3, M2 (probability that Maria wears a sweater) = 0.5

The probability that Maria eats breakfast or wears a sweater will be: M1 + M2 - M1*M2 = 0.8 - 0.15 = 0.65
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Re: The probability that Maria will eat breakfast on any given d [#permalink]
Apply the formula directly.
Both events are independent hence P (A U B ) = 0.5 * 0.3 = 0.15

P (A or B) = 0.5 + 0.3 - 0.15 = 0.65
Hence ans is B
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Re: The probability that Maria will eat breakfast on any given d [#permalink]
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Re: The probability that Maria will eat breakfast on any given d [#permalink]
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