parveen110 wrote:Mr. Smith's gardener is not dependable; the probablity that he will forget to water the rose bush during Smith's absence is 2/3. The rosebush is in questionable condition anyhow; if watered, the probability of its withering is 1/2, but if it is not watered, the probablity of its withering is 3/4. Upon returning, Smith finds that the rosebush has withered. What is the probablity that the gardener did not water the rosebush.
a. 1/2
b. 2/3
c. 1/3
d. 1/4
e. 3/4
This is an EITHER/OR group problem.
Every rosebush is either WATERED or NOT WATERED.
Every rosebush is either WITHERING or NOT WITHERING.
To organize the data, use a GROUP GRID (also known as a double-matrix).
The following fractions appear in the problem: 2/3, 1/2, and 3/4.
Let the total number of rosebushes = the LCM of the denominators = 12.
Here is the grid:
In the grid above, the entries in any row or column must add up to the TOTAL of that row or column.
The probability that he will forget to water the rose bush during Smith's absence is 2/3.
Since NOT WATERED = (2/3)(12) = 8, we get the following grid:
If watered, the probability of its withering is 1/2.
If not watered, the probability of its withering is 3/4.
Since 1/2 of the 4 watered rosebushes wither, while 3/4 of the 8 unwatered rosebushes wither, we get the following grid:
Smith finds that the rosebush has withered. What is the probablity that the gardener did not water the rosebush?
According to the grid above, (not watered and withering)/(total withering) = 6/8 = 3/4.
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