Answer:
a) 720, b) 475020, c) 69300, d) 0.146, e) 0.001
Step-by-step explanation:
It is given that my friend has 10 bottles of zinfandel, 8 of merlot, and 11 of cabernet.
a)
If he wants to serve 3 bottles of zinfandel and serving order is important, then the total number of ways is
![10\times 9\times 8=720](https://tex.z-dn.net/?f=10%5Ctimes%209%5Ctimes%208%3D720)
Therefore if he wants to serve 3 bottles of zinfandel and serving order is important, then the total number of ways are 720.
b)
The total number of bottles is
![10+8+11=29](https://tex.z-dn.net/?f=10%2B8%2B11%3D29)
Combination is defined as
![^nC_r=\frac{n!}{r!(n-r)!}](https://tex.z-dn.net/?f=%5EnC_r%3D%5Cfrac%7Bn%21%7D%7Br%21%28n-r%29%21%7D)
where, n is total possible outcomes and r is selected outcomes.
we have to select 6 bottles out of 29. so,
![^{29}C_{6}=475020](https://tex.z-dn.net/?f=%5E%7B29%7DC_%7B6%7D%3D475020)
Therefore ff 6 bottles of wine are to be randomly selected from the 29 for serving, then the total number of ways are 475020.
c)
If we want to select 2 bottles of each variety, then total number of ways are
![^{10}C_{2}\times ^{8}C_{2}\times ^{11}C_{2}=69300](https://tex.z-dn.net/?f=%5E%7B10%7DC_%7B2%7D%5Ctimes%20%5E%7B8%7DC_%7B2%7D%5Ctimes%20%5E%7B11%7DC_%7B2%7D%3D69300)
Therefore if 6 bottles are randomly selected with two bottles of each variety, then the total possible ways are 69300.
d)
Probability is defined as
![P=\frac{\text{Total outcomes}}{\text{Favorable outcomes}}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B%5Ctext%7BTotal%20outcomes%7D%7D%7B%5Ctext%7BFavorable%20outcomes%7D%7D)
![\frac{^{10}C_{2}\times ^{8}C_{2}\times ^{11}C_{2}}{^{29}C_{6}}=\frac{69300}{475020}\approx 0.146](https://tex.z-dn.net/?f=%5Cfrac%7B%5E%7B10%7DC_%7B2%7D%5Ctimes%20%5E%7B8%7DC_%7B2%7D%5Ctimes%20%5E%7B11%7DC_%7B2%7D%7D%7B%5E%7B29%7DC_%7B6%7D%7D%3D%5Cfrac%7B69300%7D%7B475020%7D%5Capprox%200.146)
Therefore the probability that two bottles of each variety being chosen is 0.146.
e)
If 6 bottles are randomly selected, then the probability that all of them are the same variety is
![\frac{^{10}C_{6}+^{8}C_{6}+^{11}C_{6}}{^{29}C_{6}}=\frac{700}{475020}\approx 0.001](https://tex.z-dn.net/?f=%5Cfrac%7B%5E%7B10%7DC_%7B6%7D%2B%5E%7B8%7DC_%7B6%7D%2B%5E%7B11%7DC_%7B6%7D%7D%7B%5E%7B29%7DC_%7B6%7D%7D%3D%5Cfrac%7B700%7D%7B475020%7D%5Capprox%200.001)
Therefore if 6 bottles are randomly selected, then the probability that all of them are the same variety is 0.001.