Answer:
a) 0.432, b) 0.1029 and c) 0.08
Step-by-step explanation:
a) what is the probability that you get 2 chocolate-flavor donuts and one vanilla-flavor donut?
Observe that there are 3 possible ways for this to happen. Getting a vanilla donut first, second or third. And every of this cases has a probability of
![\frac{6}{10}\times \frac{6}{10} \times \frac{4}{10}=\frac{144}{1000}](https://tex.z-dn.net/?f=%5Cfrac%7B6%7D%7B10%7D%5Ctimes%20%5Cfrac%7B6%7D%7B10%7D%20%5Ctimes%20%5Cfrac%7B4%7D%7B10%7D%3D%5Cfrac%7B144%7D%7B1000%7D)
Therefore, the probability of a) to happen is given by:
![3\times \frac{144}{1000}=\frac{432}{1000}=0.432](https://tex.z-dn.net/?f=3%5Ctimes%20%5Cfrac%7B144%7D%7B1000%7D%3D%5Cfrac%7B432%7D%7B1000%7D%3D0.432)
b) What is the probability that he gets 5 chocolate-flavor donuts?
Observe that:
- There are
ways to choose 5 chocolate donuts between 10 - The probability of each one of those cases to happen is given by
![(0.3)^5\times(0.7)^5](https://tex.z-dn.net/?f=%280.3%29%5E5%5Ctimes%280.7%29%5E5)
Therefore the probability of b) to happen is given by:
![{10\choose 5}\times(0.3)^5\times(0.7)^5=252\times(0.00243)\times(0.16807)\approx0.1029](https://tex.z-dn.net/?f=%7B10%5Cchoose%205%7D%5Ctimes%280.3%29%5E5%5Ctimes%280.7%29%5E5%3D252%5Ctimes%280.00243%29%5Ctimes%280.16807%29%5Capprox0.1029)
c) what is the probability that he eats 10 donuts that day?
Observe that, in order for the donut lover to eat exactly 10 donuts 2 things must happen
- He should have eaten 9 donuts from which 3 of them were chocolate-flavored.
- The 10th donut he ates must be a chocolate-flavored one.
Then we proceed to compute the probability of 1 the same way as in b).
- There are
ways to choose 3 chocolate donuts between 9 - The probability of each one of those cases to happen is given by
![(0.3)^3\times(0.7)^6](https://tex.z-dn.net/?f=%280.3%29%5E3%5Ctimes%280.7%29%5E6)
Therefore the probability of 1 to happen is given by:
![{9\choose 3}\times(0.3)^3\times(0.7)^6=84\times(0.027)\times(0.117649)\approx0.2668](https://tex.z-dn.net/?f=%7B9%5Cchoose%203%7D%5Ctimes%280.3%29%5E3%5Ctimes%280.7%29%5E6%3D84%5Ctimes%280.027%29%5Ctimes%280.117649%29%5Capprox0.2668)
Finally the probability of 2 to happen is known to be 0.3 and, as 1 and 2 are independent events, the probability of both to happen is the multiplication of the probabilities. Then, the answer for c) is:
![(0.3)\times(0.2668)\approx0.08](https://tex.z-dn.net/?f=%280.3%29%5Ctimes%280.2668%29%5Capprox0.08)