Question

Can you help me please I need help with this assignment

Answer 1

a) Consider the experiment of picking 3 tax returns at random as 3 related experiments (due to the fact that it is without replacement); then, as for the first time one selects a tax return,

$P_1(NonError)=\frac{61}{61+9}=\frac{61}{70}$

As for the second time we grab a tax return, there are 69 tax returns in total and 9 of them contain errors; thus,

$P_2(NonError)=\frac{60}{69}$

Similarly, as for the third picking round,

$P_3(NonError)=\frac{59}{68}$

Finally, the probability of experiment a) is

$\begin{gathered} P_1(NonError)*P_2(NonError)*P_3(NonError)=\frac{61*60*59}{70*69*68}=0.657471...\approx65.7\% \\ \end{gathered}$

Rounded to one decimal place, the probability of event a) is 65.7%

b) Similarly, in the event of all three tax returns containing errors,

$\begin{gathered} P_1(Error)=P_1(E)=\frac{9}{70} \\ P_2(E)=\frac{8}{69} \\ P_3(E)=\frac{7}{68} \end{gathered}$

Thus,

$\begin{gathered} \Rightarrow P(b)=P_1(E)*P_2(E)*P_3(E)=0.00153... \\ \Rightarrow P(b)\approx0.2\% \end{gathered}$

The probability of event b) is 0.2%. It is quite unusual.

c) The condition 'at least one of those containing errors' includes the cases when 1, 2, or 3 tax returns of the ones selected have errors. Notice that

$1=P(0Errors)+P(1Errors)+P(2Errors)+P(3Errors)$

Therefore,

$P(1o2o3Errors)=P(1E)+P(2E)+P(3E)=1-P(0Errors)$

And we found the probability of picking 3 tax returns without any errors in part a); thus,

$P(c)=1-0.657471...=0.342528...\approx34.3\%$

The probability of event c) is 34.3%.

d) Analogously to part c), the probability of selecting at least one without errors is

$P(d)=1-P(0NonErrors)=1-P(3Errors)=1-P(b)=0.998465...\approx99.8\%$

The probability of event d) is 99.8%, and it is not improbable at all.