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

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A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 01 May 2012, 13:34
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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
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A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 01 May 2012, 13:58
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JubtaGubar wrote:
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


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.
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Re: A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 01 May 2012, 13:36
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?
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Re: A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 01 May 2012, 14:58
Thanks Bunuel for the concise explanation!
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Re: A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 27 Mar 2014, 01:23
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JubtaGubar wrote:
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


Pie (pie) pie (cake) (cake) = (5/11) (4/10) (3/9) (6/8) (5/7) 5!/(3!.2!) = 25/77
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Re: A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 12 Nov 2016, 00:42
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JubtaGubar wrote:
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


Answer: option b

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Re: A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 19 Jun 2019, 22:00
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Bunuel wrote:
JubtaGubar wrote:
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


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;
\(C^2_5\) 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.



[b]\(C^2_5\) is # of ways to chose 2 cakes out of 6;(this should be C^2_6)?
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Re: A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 19 Jun 2019, 22:03
Williamhomey12 wrote:
Bunuel wrote:
JubtaGubar wrote:
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


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;
\(C^2_5\) 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.



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

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Re: A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 20 Jun 2019, 05:01
JubtaGubar wrote:
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


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 :cool:
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A bakery currently has 5 pies and 6 cakes in its inventory  [#permalink]

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New post 30 Jun 2019, 08:20

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

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A bakery currently has 5 pies and 6 cakes in its inventory   [#permalink] 30 Jun 2019, 08:20
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