Initially a box contained 6 blue pens, 8 black pens and 3 red pens.
You blindly picked up a pen.
(a) What is the probability that it was a blue pen?
Later some additional blue pens were put in the box.
The probability of selecting a blue pen became then 50%.
(b) How many blue pens were added to the box?
(a) p(blue) = 6/17
(b) The answer to part (b) depends on the initial conditions (which are somewhat ambiguous) at the beginning of part (b).
If the initial pen taken was black or red, then there are 6 blue and 10 non-blue pens remaining. Adding 4 blue pens will increase p(blue) to 10/20 or 50%.
If the initial pen taken was blue then there are 5 blue and 11 non-blue pens remaining. Adding 6 blue pens will increase p(blue) to 11/22 or 50%.
If the second part assumes that the random blind selection of a pen in part (a) never happened then there are 6 blue and 11 non-blue pens remaining. Adding 5 blue pens will increase p(blue) to 11/22 or 50%.
If we know a pen was taken, but don't know its color, then it is not possible to answer part (b). You could consider a weighted average: (6/17)*6 + (11/17)*4 = 80/17, but I doubt this is the answer being sought.
So the answer to part (b) could be either 4, 5, 6, 80/17, or impossible to answer, depending on how the problem is interpreted.
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Posted by Larry
on 2024-03-18 08:58:22 |
different interpretations
