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The probability that event A occurs is 0.5 and the probability that ev

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The probability that event A occurs is 0.5 and the probability that ev  [#permalink]

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New post 28 Aug 2017, 23:56
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The probability that event A occurs is 0.5 and the probability that event B occurs is 0.7. What is the probability that event A occurs but not event B?

1) Event A and event B are independent.
2) The probability that neither event A nor event B occurs is 0.15.

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Re: The probability that event A occurs is 0.5 and the probability that ev  [#permalink]

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New post 29 Aug 2017, 07:32
MathRevolution wrote:
The probability that event A occurs is 0.5 and the probability that event B occurs is 0.7. What is the probability that event A occurs but not event B?

1) Event A and event B are independent.
2) The probability that neither event A nor event B occurs is 0.15.


lets see the statements ..

1) Event A and event B are independent.
so P is P(A)*{1-P(B)}=0.5*(1-0.7)=0.15..
suff

2) The probability that neither event A nor event B occurs is 0.15
now NONE = 0.15...
but it is same as (1-P(A))(1-P(B))=(1-0.5)(1-0.7)=0.5*0.3=0.15..
this means the events are independent..
again suff

D
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Re: The probability that event A occurs is 0.5 and the probability that ev  [#permalink]

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New post 30 Aug 2017, 01:25
=>Condition 1)
We call an event A and an event B independent if P(A∩B) = P(A)*P(B).
P(A∩B) = P(A)*P(B) = 0.5 * 0.7 = 0.35.
P(A∩BC) = P(A) - P(A∩B) = 0.5 – 0.35 = 0.15.
This is sufficient.

Condition 2)
1 – P(A∪B) = 0.15. It means P(A∪B) = 0.85.
P(A∪B) = P(A) + P(B) - P(A∩B) = 0.5 + 0.7 - P(A∩B) = 0.85.
Thus P(A∩B) = 0.15.
P(A∩B^C) = P(A) - P(A∩B) = 0.5 – 0.35 = 0.15.
This is also sufficient too.

Ans: D
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Re: The probability that event A occurs is 0.5 and the probability that ev &nbs [#permalink] 30 Aug 2017, 01:25
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