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.
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80% (02:42) correct
