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Repeated outcomes questions

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Repeated outcomes questions

by hjafferi » Sun Jun 23, 2013 9:22 am
Qs. 1 : Ann is making fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in hte basket as long as there are some fruits in the basket. how many different choices does she have? (assuming the furits are all different from each other)

Qs. 2 : Ann is making fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in hte basket as long as there is atleast one of each kind. how many different choices does she have? (assuming the furits are all different from each other)

Qs. 3 there are three secretaries who work for three departments. if each of the three departments have one report to be typed out and the reports are randomly assigned to a secretary. what is the probability that all three secretaries are assigned atleast one report?

Qs. 4 There are three secretaries who work for four departments. If each of the four departments have one report to be typed out. and the reports are randomly assigned toa secretary . what is the probablity that all three secretaries are assigned atleast one report?
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by GMATGuruNY » Sun Jun 23, 2013 12:55 pm
hjafferi wrote: Qs. 3 there are three secretaries who work for three departments. if each of the three departments have one report to be typed out and the reports are randomly assigned to a secretary. what is the probability that all three secretaries are assigned atleast one report?
P = (good outcomes)/(total possible outcomes).

Let the 3 reports be A, B, and C.
Each report must be assigned to a secretary.

Total possible outcomes:
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 3. (Any of the 3 secretaries.)
Number of options for C = 3. (Any of the 3 secretaries.)
To combine these options, we multiply:
3*3*3 = 27.

Good outcomes:
A good outcome occurs when each report is assigned to a DIFFERENT secretary.
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 2. (Either of the 2 remaining secretaries.)
Number of options for C = 1. (Only 1 secretary left.).
To combined these options, we multiply:
3*2*1 = 6.

Thus:
(good outcomes)/(total possible outcomes) = 6/27 = 2/9.
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by GMATGuruNY » Sun Jun 23, 2013 1:09 pm
hjafferi wrote: Qs. 4 There are three secretaries who work for four departments. If each of the four departments have one report to be typed out. and the reports are randomly assigned toa secretary . what is the probablity that all three secretaries are assigned atleast one report?
P = (good outcomes)/(total possible outcomes).

Let the 4 reports be A, B, C and D.
Each report must be assigned to a secretary.

Total possible outcomes:
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 3. (Any of the 3 secretaries.)
Number of options for C = 3. (Any of the 3 secretaries.)
Number of options of D = 3. (Any of the 3 secretaries.)
To combine these options, we multiply:
3*3*3*3 = 81.

Good outcomes:
For each secretary to be assigned at least 1 report, exactly 1 secretary must receive a PAIR of reports, while the other 2 secretaries receive 1 report each.
Number of pairs that can be formed from the 4 reports = 4C2 = (4*3)/(2*1) = 6.
Number of ways to assign this pair = 3. (Any of the 3 secretaries.)
Number of ways to assign the next report = 2. (Either of the 2 remaining secretaries.)
Number of ways to assign the last report = 1. (Only 1 secretary left.)
To combine these options, we multiply:
6*3*2*1 = 36.

Thus:
(good outcomes)/(total possible outcomes) = 36/81= 4/9.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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I unlock the best way for YOU to solve problems.

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