Question

(1) In a bag of 10 donuts: every donut looks the same but 6 of them have the chocolate flavor and the other 4 have the vanilla flavor. Now you pick 3 donuts from these 10 donuts, what is the probability that you get 2 chocolate-flavor donuts and one vanilla-flavor donut? (2) A donut lover works in a donut factory and he works on a particular production line that produces chocolate-flavor and vanilla-flavor donuts. He hears from his boss that 30% of the donuts in that production line are chocolate-flavor ones. If one day, he will not resist so he will eat 10 donuts when he is working on that production line. What is the probability that he gets 5 chocolate-flavor donuts? (3) The same guy from question (2): If one day, he decides that he need to eat 4 chocolate-flavor donuts until he stops eating. So what is the probability that he eats 10 donuts that day?

Answer 1

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

$\frac{6}{10}\times \frac{6}{10} \times \frac{4}{10}=\frac{144}{1000}$

Therefore, the probability of a) to happen is given by:

$3\times \frac{144}{1000}=\frac{432}{1000}=0.432$

b) What is the probability that he gets 5 chocolate-flavor donuts?

Observe that:

• There are ${10 \choose 5}$ ways to choose 5 chocolate donuts between 10
• The probability of each one of those cases to happen is given by $(0.3)^5\times(0.7)^5$

Therefore the probability of b) to happen is given by:

${10\choose 5}\times(0.3)^5\times(0.7)^5=252\times(0.00243)\times(0.16807)\approx0.1029$

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

1. He should have eaten 9 donuts from which 3 of them were chocolate-flavored.
2. 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 ${9 \choose 3}$ ways to choose 3 chocolate donuts between 9
• The probability of each one of those cases to happen is given by $(0.3)^3\times(0.7)^6$

Therefore the probability of 1 to happen is given by:

${9\choose 3}\times(0.3)^3\times(0.7)^6=84\times(0.027)\times(0.117649)\approx0.2668$

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:

$(0.3)\times(0.2668)\approx0.08$