Find the requested probability.
P(A) if P(A) = 0.6. the first A has a line over it. than the P (A) line over it
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p(NOT A)=1-P(A)
P(mean of A) if mean P(A) = .6
If we have six Apples and four Oranges in a box
then P(A) = .6 when we draw one thing at random.
If we have a whole bunch of boxes and on the average there are six apples in each then P(A) = .6
BUT
The probability that we will draw that mean on any draw depends on how the apples are distributed around six in the boxes.
If all the boxes had 6 apples, then P(of the mean of 6) would be 1
but if half had seven and half had five, P(mean A) would be zero. In other words I think the answer depends on your distribution and its standard deviation from the mean.
The notation "P(A)" with a line over it typically represents the complement of event A. The complement of an event refers to the probability of the event not happening. Therefore, if P(A) = 0.6, then the complement of event A, denoted as P(A̅), would be 1 - P(A) = 1 - 0.6 = 0.4.
To find the requested probability for P(A̅), where A̅ has a line over it, we need to understand what it represents.
When a line is placed over a variable, it typically denotes the complement of that variable. In probability theory, the complement of an event A is the event that A does not occur.
So, P(A̅) represents the probability that event A does not occur. In other words, it is the probability of the event not happening.
Given that P(A) = 0.6, we can find P(A̅) by subtracting P(A) from 1 since the sum of the probabilities of an event and its complement must equal 1.
P(A̅) = 1 - P(A) = 1 - 0.6 = 0.4
Therefore, the requested probability of P(A̅) is 0.4.