Question

Shankar has decided to train to be a Carbucks Barrista. Being young and inexperienced, for every order he makes a mistake in making that order with probability 1/3 and makes the order correctly with probability 2/3, with the probabilities of making an error independent across different orders.
a. Shankar comes into work Monday morning. What is the probability that he makes no mistakes on his first 10 orders but the 11th order is a mistake?
b. Another Employee (Fran) and Shankar decide to have a competition: Every customer that comes in, both Fran and Shankar would make the order for that person (so each person would get 2 of the same item!). The first amongst either Fran or Shankar that makes a mistake quits Carbucks and goes to grad school to learn probability.
If Fran is more experienced and makes mistakes on an order with probability 1/6 independent across orders and independent of what Shankar is doing on an order, what is the probability that Shankar quits and goes to grad school?

Answer 1

Answer:

Step-by-step explanation:

Since each trial is independent of the other

no of mistakes he does is binomial with p = 1/3

a) the probability that he makes no mistakes on his first 10 orders but the 11th order is a mistake

= $(\frac{2}{3}) ^{10} *\frac{1}{3}\\=\frac{2^{10} }{3^{11} }$

b) Prob that shanker quits = P(Shankar does I one mistake and Fran does not do the first one)+Prob (Shanker does mistake in the II one while Fran does both right)

= $\frac{1*5}{3*6} +\frac{2}{3} \frac{1}{3}(\frac{5}{6})^2 =\frac{5}{18} +\frac{50}{216} \\=\frac{55}{108}$