Independent Probabilities

[jsbeckton]

70% of all vehicles at an inspection plant pass inspection. Assuming successive vehicles pass or fail independently of one another, calculate the following:

a) P(all of the next 3 vehicles pass) anwser =0.343

\(\displaystyle \frac{7}{{10}} \times \frac{7}{{10}} \times \frac{7}{{10}} = 0.343\)


b) P(at least one of the next 3 fails) anwser = 0.657

\(\displaystyle \begin{array}{l}
\frac{{30vehicles}}{{29pass}} \times \frac{3}{{10}} = .290 \\
\frac{{28vehicles}}{{30pass}} \times \frac{3}{{10}} = .280 \\
\frac{{27vehicles}}{{30pass}} \times \frac{3}{{10}} = .270 \\
\\
Total = .840 \\
\end{array}\)


c) P(exactly one of the next 3 passes) anwser = 0.189


d) P(at most, one of the next 3 passes) anwser = 0.216

e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)? anwser = 0.353

I've figured the first part out but obviously my logic stops there, can anyone point me in the right direction here? Thanks.

 

[galactus]

You can use the binomial distribution.

\(\displaystyle p^{3}+3p^{2}q+3pq^{2}+q^{3}\)

a). is correct.

b). \(\displaystyle 1-.343=.657\)

c). \(\displaystyle 3pq^{2}=3(.7)(.3)^{2}=.189\)

d). \(\displaystyle 3pq^{2}+q^{3}=3(.7)(.3)^{2}+(.3)^{3}=.216\)

e). \(\displaystyle p^{3}+3p^{2}q+3pq^{2}=(.7)^{3}+3(.7)^{2}(.3)+3(.7)(.3)^{2}=.973\); \(\displaystyle \frac{.343}{.973}=.353\)

 

[jsbeckton]

thanks! How in the world could I not see (b), I guess I was thinking way outside the box.