Thread: Basic Probability

The probability a student will pass Test 1 is 4/5.
The probability that he will pass Test 2 is 3/4.
The probability that he will pass both tests is 2/3.


A.) What is the probability that he passes at least one test?

B.) What is the probability that he fails both tests?



A.) (4/5) + (3/4) = (31/20) - (12/20) = 19/20

B.) (1/5) x (1/4) = 1/20

Please correct me if I'm wrong, thanks!

Originally Posted by toop

The probability a student will pass Test 1 is 4/5.
The probability that he will pass Test 2 is 3/4.
The probability that he will pass both tests is 2/3.A.) What is the probability that he passes at least one test?

B.) What is the probability that he fails both tests?



A.) (4/5) + (3/4) = (31/20) - (12/20) = 19/20

B.) (1/5) x (1/4) = 1/20

Please correct me if I'm wrong, thanks!

Your calculations have assumed that the passing test 1 and passing test 2 are independent events. They are not: $\displaystyle \frac{4}{5} \cdot \frac{3}{4} \neq \frac{2}{3}$. So your answers are wrong.

A simple approach is to make a Karnaugh table:



$\displaystyle \begin{tabular}{l | c | c | c} & Pass test 1 & Fail test 1 & \\ \hline Pass test 2 & 2/3 & & 3/4\\ \hline Fail test 2 & & & \\ \hline & 4/5 & & 1 \\ \end{tabular}
$



$\displaystyle \begin{tabular}{l | c | c | c} & Pass test 1 & Fail test 1 & \\ \hline Pass test 2 & 2/3 & a & 3/4\\ \hline Fail test 2 & b & c & d \\ \hline & 4/5 & e & 1 \\ \end{tabular}
$


Note that:


$\displaystyle \frac{2}{3} + a = \frac{3}{4} \Rightarrow a = \frac{1}{12}$.

$\displaystyle \frac{2}{3} + b = \frac{4}{5} \Rightarrow b = \frac{2}{15}$.

$\displaystyle \frac{3}{4} + d = 1 \Rightarrow d = \frac{1}{4}$.

$\displaystyle \frac{4}{5} + e = 1 \Rightarrow e = \frac{1}{5}$.


I will leave $\displaystyle c$ for you to calculate. Then:


Pr(fails both tests) = c.

Pr(passes at least one test) = 1 - Pr(fails both tests) = 1 - c.

Pr(fail both tests) = 7/60

Pr(pass at least one) = 53/60


I understood how the answer was derived, but is there another way to do it besides using the Karnaugh table?

Thanks

Originally Posted by toop

Pr(fail both tests) = 7/60

Pr(pass at least one) = 53/60


I understood how the answer was derived, but is there another way to do it besides using the Karnaugh table?

Thanks

There are several ways. It can certainly be done using the various probability formulae you've learned.

But why do something the hard way when you can do it the easy way ....?

Plus, I always struggle with formulae and prefer a visual approach where possible. Someone else might like to show a formula based approach - I'd just make a fool of myself.

(a) $\displaystyle P(A \cup B)=P(A)+P(B)-P(AB)= {4\over 5}+{3\over 4}-{2\over 3} $.

(b) 1- answer in (a).