At least one… not
all...?
I often saw that some students have a problem understanding
questions that involve “at least one” or “not all”.
The first time that this really becomes an issue is when independence it
introduced. Here is how I graphically explain that concept:
Let’s assume four independent events:
P(Andy graduates within 4 years) = P(A) = 60%
P(Beth falls asleep in class) = P(B) = 20%
P(Chris will have sunshine during his wedding this summer) =
P(C) = 70%
P(Derrick will beat me in racquetball tonight) = P(D) = 10%
Questions that often cause trouble are:
“What is the chance that
not all of these events happen?”
which is the same as
“What is the chance at
most 3 events happen?”
or
“What is the chance at
least one of these events happen?”
When my students come up with all kinds of crazy answers of
how to answer this, I show them the following graph:
By virtue of independence, P(ALL) = P(A)*P(B)*P(C)*P(D) =
0.6*0.2*0.7*0.1 = 0.84%
and since ALL and NOT ALL are complements(not opposites!),
P(NOT ALL) = 100% - P(ALL) = 99.16%
Similarly P(NONE) = P(A̅)*P(B̅)*P(C̅)*P(D̅) = (1-0.6)*(1-0.2)*(1-0.7)*(1-0.1)
= 8.64%
and since NONE and AT LEAST ONE are complements, P(AT LEAST
ONE) = 100% - P(NONE) = 91.36%
If students don’t believe that the rule of complements (P(A)
= 1-P(A̅) is useful, it is time to show them a four-event Venn diagram:
The first one just shows the four Events: A, B, C, D.
The second shows for all the 16 combinations of events happening and
not happening how many actually happen.
Let’s go back to P(ALL) which means A, B, C and D must
happen – that are is indicated with the “4” in the right diagram . Then P(NOT
ALL) is everything else – all 15 combinations of events happening and not
happening (Those areas indicated with a 0, 1, 2 or 3). Since they are all mutually
exclusive outcomes, nothing prevents us from computing their respective
probabilities, but if asked P(NOT ALL), it is much easier to acknowledge that
P(NOT ALL) = 100%-P(ALL)
The same logic holds for P(AT LEAST ONE). At least one event
happening (so 1, 2, 3 or 4) is everything but the outside of the eclipes, so again there would
be 15 different combinations of events happening and not happening to be accounted
for. Instead it is much more time efficient to compute
P(AT LEAST ONE) = 100%-P(NONE)
