The probability to fail a drive test is $t$ while the probability to pass a drive test is $t+0.2$. I really need to know if what I am doing is correct. I guess I should be using bernoulli but I don't know how.
a. What is the probability to pass?
Sol: $0.6$.
b. Let Donna, David, Johnny and Natalie be four random people. What is the probability that exactly two people pass the test?
Sol: $4\cdot(0.6\cdot 0.6\cdot 0.4 \cdot 0.4)$ (Of that one I am really not sure).
c. Given only two of the four has passed the test, what is the probability those were Donna and Johnny?
Sol: Every one of them has either failed or passed where the chance is $50\%$. The probability it is Johnny is $0.5$, while the probability it is Johnny and Donna is $0.5\cdot 0.5$.
d. Is the probability that at least one of the four passes the test greater than the probability that at least one of the four fails?
That one I don't remember how to solve. I can do it using discrete mathematics but that is not the point. I would appreciate your help.
Best Answer
(a) is correct
(b), as the comments say, should be $\binom{4}{2} 0.6^2 0.4^2$. $\binom{4}{2} = 6$ is the number of possible pairs candidates who pass, i.e. $\{Donna,David\}$,$\{Donna,Johnny\}$,$\{Donna,Natalie\}$,$\{David,Johnny\}$,$\{David,Natalie\}$ or $\{Johnny,Natalie\}$.
(c) You already know how many people have passed, the questions is if it where exactly Donna and Johnny. The above shows that there are 6 possible ways in which exactly two people can pass, the probability of any specific pair is thus $\frac{1}{6}$.
(d) Compute it. The probability that at least one passes is the same as the probability that not all four fail. The probability of that is $\binom{4}{0} 0.4^4 0.6^0 = \frac{256}{10000} = 0.0256$, meaning the probability that at least one candidate passes is $1-0.0256$ = $0.9764$.
The probability that at least one candidate fails is the same as the probability that not all four pass, which is $\binom{4}{4} 0.4^0 0.6^4 = 0.1296$. The probability of at least one candidate failing is thus $0.8704$.
The answer is thus yes, the probabilit of one candidate passing is greater than the probability of at least one failing.