There is an 80% chance that Deeksha will skip her lunch and 25% chanc

Question

There is an 80% chance that Deeksha will skip her lunch and 25% chance that there will be a power failure. If these events are independent, what is the probability that Deeksha will skip her lunch OR that there will be a power failure?

(A) 20%
(B) 80%
(C) 85%
(D) 95%
(E) 105%

(A) 20%
(B) 80%
(C) 85%
(D) 95%
(E) 105%

Abhishek009 · Answer

100-\frac{(100-80)(100-25)}{100}

= 100 - 15

= 85 , Answer must be (C)

sekipap · Answer

basically, the sum of the two events, minus the interception (which would be double counted if we dont erase it)
so, interception is 80%*25% = 20%
the sum is 105% --> result = 105-20 = 85%

Fdambro294 · Answer

We can solve this problem because the events are Independent, thought NOT mutually exclusive

P (skip OR failure) = P(skip) + P(failure) - P(skip AND failure)

For independent events, the probability of 2 unique, independent events occurring is simply the PRODUCT of the 2 individual probabilities

P(skip AND failure) = P(skip) * P(failure) = (.8) (.25) = .200

Answer:

(.8) + (.25) - (.2) = .85

Or

85%

Posted from my mobile device

gmatstuff123 · Answer

Little confused?
Since they are independent, why can't we just add the probability?

bumpbot · Answer

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There is an 80% chance that Deeksha will skip her lunch and 25% chanc

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There is an 80% chance that Deeksha will skip her lunch and 25% chanc

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