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A bakery currently has 5 pies and 6 cakes in its inventory 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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B
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A bakery currently has 5 pies and 6 cakes in its inventory 01 May 2012, 13:58
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JubtaGubar
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
Show more

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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JubtaGubar
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A bakery currently has 5 pies and 6 cakes in its inventory 01 May 2012, 13:36
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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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A bakery currently has 5 pies and 6 cakes in its inventory 01 May 2012, 14:58
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Thanks Bunuel for the concise explanation!
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A bakery currently has 5 pies and 6 cakes in its inventory 27 Mar 2014, 01:23
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JubtaGubar
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
Show more

Pie (pie) pie (cake) (cake) = (5/11) (4/10) (3/9) (6/8) (5/7) 5!/(3!.2!) = 25/77
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A bakery currently has 5 pies and 6 cakes in its inventory 12 Nov 2016, 00:42
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JubtaGubar
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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Answer: option b

Check solution
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A bakery currently has 5 pies and 6 cakes in its inventory 19 Jun 2019, 22:00
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Bunuel
JubtaGubar
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.
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[b]\(C^2_5\) is # of ways to chose 2 cakes out of 6;(this should be C^2_6)?
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A bakery currently has 5 pies and 6 cakes in its inventory 19 Jun 2019, 22:03
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JubtaGubar
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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Edited the typo. Thank you.
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A bakery currently has 5 pies and 6 cakes in its inventory 20 Jun 2019, 05:01
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JubtaGubar
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
Show more

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 30 Jun 2019, 08:20
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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 21 Jul 2020, 16:47
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Quote:
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?
Show more

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

Probability = 150/462 -> 25/77

Answer -> B
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A bakery currently has 5 pies and 6 cakes in its inventory 22 Jul 2020, 14:20
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JubtaGubar
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?
Show more



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 \(\frac{5}{77}\) * \(\frac{5!}{3!*2!}\)
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A bakery currently has 5 pies and 6 cakes in its inventory 06 Dec 2020, 17:14
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you can just do 1 / (3/5/11x6).
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A bakery currently has 5 pies and 6 cakes in its inventory 18 Jan 2026, 20:50
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