SOLUTION: Given that a box contains 12 spiders, of which 3 are tarantulas, 4 black widows, and 5 jumping spiders. If we choose 2 with replacement, what is the probability of getting one bla

Question 477724: Given that a box contains 12 spiders, of
which 3 are tarantulas, 4 black widows, and 5 jumping spiders. If we choose 2 with replacement, what is the probability of getting one black widow and one that is not a black widow. If we choose them without replacement, the probabilities would change from the first selection to the second. But choosing with replacement means the probabilities don�t change from the first to
the second selection. Then clearly the two events (first choice and second choice) are independent events.
Now the probability of choosing a black widow is P(B) = 4/12 = 1/3 since there are 4 black
widows in the 12 spiders.
And the probability of choosing one that is not a black widow is P(B) = 8/12 = 2/3.
Since these events are independent, we could use the formula for independent events, which is
P(B and B) = P(B) � P(B) = 1/3 � 2/3 = 2/9 = 22.2%.
The reasoning seems correct, the only problem with this is that it gives the wrong answer! For this
assignment, explain in detail why the formula, in this case, gives an answer that is not correct.
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
There are actually two issues arising here, namely, INDEPENDENCE and MUTUAL EXCLUSIVITY.
It seems that you are identifying idependence with mutual exclusivity, which are two different things.
The events B and B' are mutually exclusive (or disjoint), but they are not independent.
Hence P(B and B') = P(empty set) = 0, since B and B' are complementary events.
If two events are mutually exclusive, then the correct rule to use is
P(B or B') = P(B) + P(B') = 1/3 + 2/3 = 1.
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