Answer :
Answer:
a) 336
b) 593775
c) 83160
d) P=0.14
e) P=0.0019
Step-by-step explanation:
We have wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet.
a) If he wants to serve 3 bottles of zinfandel and serving order is important. We get:
C=8·7·6=336
b) {30}_C_{6}=\frac{30!}{6!(30-6)!}
{30}_C_{6}=593775
c) {8}_C_{2} · {10}_C_{2} · {12}_C_{2}=
=\frac{8!}{2!(8-2)!} · \frac{10!}{2!(10-2)!} · \frac{12!}{2!(12-2)!}
=28 · 45 · 66
=83160
d) We calculate the number of possible combinations:
{30}_C_{6}=593775
We calculate the number of favorable combinations:
{8}_C_{2} · {10}_C_{2} · {12}_C_{2}=83160
The probability that this results in two bottles of each variety being is
P=83160/593775
P=0.14
e) We calculate the number of possible combinations:
{30}_C_{6}=593775
We calculate the number of favorable combinations:
{8}_C_{6} + {10}_C_{6} + {12}_C_{6}= 28+210+924=1162
The probability is
P=1162/593775
P=0.0019
Final answer:
A friend has several options for serving wine at a dinner party, including 336 ways to serve 3 bottles of zinfandel in order, 593,775 ways to randomly select 6 bottles from 30, and 90 ways to select 2 bottles of each variety among 6 bottles. The probabilities for the specific selections are 0.01515% and 0.11707%, respectively.
Explanation:
Your friend has various options for serving wine at a dinner party:
If he wants to serve 3 bottles of zinfandel and the serving order is important, he has Permutation options: 8P3 = 8! / (8-3)! = 8 x 7 x 6 = 336 ways to do this.
If 6 bottles of wine are randomly selected from the total of 30, he has Combination options: 30C6 = 30! / (6! (30-6)!) = 593,775 ways to do this.
If 6 bottles are randomly selected and the goal is to have two bottles of each variety, we use the formula for multinomial coefficients: (6!)/(2!2!2!) = 90 ways to achieve this.
The probability of selecting exactly two bottles of each variety is the number of favorable outcomes divided by the total number of outcomes: (90/593,775) = 0.0001515, or about 0.01515%.
The probability that all 6 randomly selected bottles are of the same variety can be calculated by adding the probabilities for each variety: (8C6 + 10C6 + 12C6) / 30C6 = (28 + 210 + 924) / 593,775 = 0.0011707, or about 0.11707%.
