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
= [tex](\frac{2}{3}) ^{10} *\frac{1}{3}\\=\frac{2^{10} }{3^{11} }[/tex]
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)
= [tex]\frac{1*5}{3*6} +\frac{2}{3} \frac{1}{3}(\frac{5}{6})^2 =\frac{5}{18} +\frac{50}{216} \\=\frac{55}{108}[/tex]