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I solved that in another way:

Probability of both p^2
Probability of no E and no F (1-p)^2
as p is equal on both cases

All - both - neither
---------------
2

(1 - (1-p)^2 - p^2)/2 = 2(p-p^2)/2
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Why it cant be D p^2-p Bunuel
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MalachiKeti
Why it cant be D p^2-p Bunuel

How are you getting p^2 - p? What’s your reasoning behind it? I think if you think it through carefully, you’ll spot the flaw in your reasoning yourself.
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This problem was easier for me to plug and check. You would want to pick an easy number to understand, so in this case, let's say each event has a 10% or 1/10 chance of happening.

That means that odds of the event described will be (1/10)*(9/10) which is 9/100. Now we plug and go

(1*10)^2 Nope
1-(1*10)^2 Nope
ETC
Until E
(1/10)-(1/10)^2

(1/10)-(1/100)

(10/100)-(1/100)= 9/100

Answer Choice E

rakman123
If events E and F are independent and each has a probability p of occurring, which of the following represents the probability that E will occur but F will not occur?

A) p^2
B) 1-p^2
C) p^2-1
D) p^2-p
E) p-p^2
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Probability of occurance of event = p , so probability of event not occuring is 1-p
P(E Occur) and P (F not occur)=
p (1-p)
p-p^2
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