Question 1174649: A teacher has two boxes of pens. Box 1 contains 7 blue pens and 3 black
pens, box 2 contains 2 blue pens and 8 black pens. The probability of
selecting a pen from box 1 is 2
3
and the probability of selecting a pen
from box 2 is 1
3
. If a pen is selected at random what is the probability
that it is from box 1 given that it is black?
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! P (box 1|black)=P(Black|box 1)*Box 1/P(black)
=0.3*(2/3)/(14/30), the expected probability of a black pen's being chosen with the above probabilities.
=2/10/14/30 or 2*3/14
=0.429.
another way
-
=====BLU- BLK-TOT
Box 1--7----3---10
Box 1--7----3---10
Box 2-2----8---10]
twice as many box 1 because 2/3 probability
this leads to 16 blue an 14 black for a single choice which is where the 14/30 probability for a black comes from.
Then given that a black pen is chosen (6/14), six came from box 1, so the probability is 0.429, equalling the one above.
Answer by ikleyn(52767)  (Show Source):
You can put this solution on YOUR website! .
A teacher has two boxes of pens. Box 1 contains 7 blue pens and 3 black
pens, box 2 contains 2 blue pens and 8 black pens. The probability of
selecting a pen from box 1 is 2
3
and the probability of selecting a pen
from box 2 is 1
3
. If a pen is selected at random what is the probability
that it is from box 1 given that it is black?
~~~~~~~~~~~~~
As the post is worded, printed, posted and presented, it MAKES no SENSE.
Indeed, this passage from the post "The probability of selecting a pen from box 1 is 2/3" is NONSENSICAL
without pointing to which color pen it does relate.
Same the next passage "the probability of selecting a pen from box 2 is 1/3" is NONSENSICAL due to the same reason.
Since the post makes no sense, the solution makes no sense, too.
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