A bakery currently has 5 pies and 6 cakes in its inventory

Question

A bakery currently has 5 pies and 6 cakes in its inventory. The bakery’s owner has decided to display 5 of these items in the bakery’s front window. If the items are randomly selected, what is the probability that the display will have exactly 3 pies?

A 3/11
B. 25/77
C. 5/11
D. 7/11
E. 50/77

A 3/11
B. 25/77
C. 5/11
D. 7/11
E. 50/77

Bunuel · Accepted Answer

Combination approach:

\frac{C^3_5*C^2_6}{C^5_{11}}=\frac{25}{77} , where:
 C^3_5  is # of ways to chose 3 pies out of 5;
 C^2_6  is # of ways to chose 2 cakes out of 6;
 C^5_{11}  is # of ways to chose 5 items out of total 5+6=11.

Answer: B.

Probability approach:

We need the probability that the owner will select PPPCC:  \frac{5!}{3!*2!}*(\frac{5}{11}*\frac{4}{10}*\frac{3}{9})*(\frac{6}{8}*\frac{5}{7})=\frac{25}{77} , we are multiplying by  \frac{5!}{3!*2!}  since PPPCC scenario can occur in number of ways: PPPCC, PPCPC, PCPPC, CPPPC, ... Notice that the number of occurrences of PPPCC basically is the number of arrangements of 5 letters PPPCC out of which 3 P's and 2 C's are identical, so it's  \frac{5!}{3!*2!}  .

Answer: B.

Hope it's clear.

JubtaGubar · Answer

My calculation is the following:
5/11 * 4/10 * 3/9 * 6/8 * 5/7 = 5/77
but there is no such solution. 
could you please explain why is this method incorrect?

Zynga · Answer

Thanks Bunuel for the concise explanation!

PerfectScores · Answer

Pie (pie) pie (cake) (cake) = (5/11) (4/10) (3/9) (6/8) (5/7) 5!/(3!.2!) = 25/77

GMATinsight · Answer

Answer: option b

Check solution

Williamhomey12 · Answer

C^2_5  is # of ways to chose 2 cakes out of 6;(this should be C^2_6)?

Bunuel · Answer

_________________________
Edited the typo. Thank you.

Archit3110 · Answer

total items; 11 so arragement of 5 items at display can be done in 11c5 ways ; and exactly 3 pies and 2 cakes; 5c3*6c2 so P = 5c3*6c2/ 11c5 = 25/77 IMO B

EgmatQuantExpert · Answer

Solution 
  Given:  
 • A bakery currently has 5 pies and 6 cakes in its inventory
• The bakery’s owner has decided to display 5 of these items in the bakery’s front window

To find:  
 • The probability that the display will have exactly 3 pies

Approach and Working Out:   
 • Sample size =  ^{11}C_5 
• Number of ways of selecting 5 items such that the display will have exactly 3 pies =  ^5C_3 * ^6C_2 • Therefore, the probability =  \frac{^5C_3 * ^6C_2}{^{11}C_5} = \frac{25}{77}

Hence, the correct answer is Option B.

Answer: B

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-AT-e1533292152363.png

daniformic · Answer

Total outcomes = 11!/(5!*6!) -> 462
Desired outcomes = 5!/(3!*2!) * 6!/(2!*4!) -> 150

Probability = 150/462 -> 25/77

Answer -> B

Swat2020 · Answer

That is because you have only considered the situation where three pies are selected first and then two cakes are selected, i.e PPPCC.
However. you need to take into account other scenarios also, such as PCCPP, CCPPP etc.
Therefore solution will be  \frac{5}{77}  *  \frac{5!}{3!*2!}

Lagasarous · Answer

you can just do 1 / (3/5/11x6).

bumpbot · Answer

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A bakery currently has 5 pies and 6 cakes in its inventory

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