- #1
Townsend
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If event S has a 40 percent chance of happening on the first try and a 65 percent chance of happening on the second try what is the chance of event A happening?
Let A be the event S happens on the first try, note [tex]A^c[/tex] is the complement of A.
Let B be the event S happens on the second try.
I reasoned it to be a 79 percent chance of S happening.
1. A and B are mutually exclusive, in other words if A happens B does not, vise versa.
2. The probability of event B happening depends on event A not happening.
Thus [tex]P(B \cap A^c) = .39[/tex], so P(S) = P(A) + [tex]P(B \cap A^c)[/tex] = .79.
So I reason a 79 percent chance of event S.
Thanks
Let A be the event S happens on the first try, note [tex]A^c[/tex] is the complement of A.
Let B be the event S happens on the second try.
I reasoned it to be a 79 percent chance of S happening.
1. A and B are mutually exclusive, in other words if A happens B does not, vise versa.
2. The probability of event B happening depends on event A not happening.
Thus [tex]P(B \cap A^c) = .39[/tex], so P(S) = P(A) + [tex]P(B \cap A^c)[/tex] = .79.
So I reason a 79 percent chance of event S.
Thanks