0
$\begingroup$

Each of them has 3 answers - only one is correct. In order to pass the test, a student has to answer at least 4 questions correctly. Alan didn't study and guessed all of the answers. What is the chance of Alan failing the test?

My idea is: $ \left(\frac{2}{3}\right)^3 + \left(\frac{2}{3}\right)^4 + \left(\frac{2}{3}\right)^5 + \left(\frac{2}{3}\right)^6$

because there is a $\frac{2}{3}$ chance of an answer being wrong and it is exponentiated for 3, 4, 5 and 6 wrong questions (which you need for failing the test). Is this correct?

$\endgroup$
2

2 Answers 2

4
$\begingroup$

As said in other comments/answers, binomial distribution works a charm.

However, if this is too complex for you, we can do the old complementary probability with cases method.

The probability of getting a question right is $\frac{1}{3}$ and wrong is $\frac{2}{3}$. It's easier to calculate the chance that Alan passes the test, then subtract that probability from $1$.

We can split this into 3 cases:

Alan gets 4 correct

The probability of this is $(\frac{1}{3})^4 \cdot (\frac{2}{3})^2 \cdot \binom{6}{2}.$ The reason we add $\binom{6}{2}$, or $6$ choose $2$, is because we can order the correct and wrong answers in $\binom{6}{2}$ ways. For example, Alan can get CCCCII (where C is correct and I is incorrect), but he could also get CCCICI. There are $\binom{6}{2}$ ways to order this; this is something you must be familiar and understand well to know the gist of probaiblity.

Anyway, $(\frac{1}{3})^4 \cdot (\frac{2}{3})^2 \cdot \binom{6}{2} \approx 0.082304.$

From here, I hope you can get the rest. We have to compute the Alan gets 5 correct and Alan gets 6 correct cases, then add everything up and remember to subtract by 1 at the end.

Additionally, the reason your answer is wrong is because the probabilities are not part of different results; they from the same result. You would have to multiply those individual probabilities to get closer to the answer, but then you would be missing the choose functions. This method I've written (with cases and complementary probability) is generally more reliable, albeit a bit tedious. (Binomial distribution works too and is very efficient for this problem, but as said if you don't understand it you can fall back on this reliable method.)

Hope that helped.

-FruDe

$\endgroup$
2
  • $\begingroup$ Thank you so much! $\endgroup$ Commented Jun 8, 2020 at 13:06
  • $\begingroup$ @JergušBobák Yeah, sure thing. Hopefully you understand everything I explained. $\endgroup$ Commented Jun 8, 2020 at 13:32
0
$\begingroup$

Hint : Binomially expand ... \begin{eqnarray*} \left( \frac{1}{3}+\frac{2}{3} \right)^6= \color{red}{\left( \frac{2}{3} \right)^6 +6 \left( \frac{2}{3}\right)^5 \left( \frac{1}{3} \right)^1 +15 \left( \frac{2}{3}\right)^4 \left( \frac{1}{3} \right)^2+20 \left( \frac{2}{3}\right)^3 \left( \frac{1}{3} \right)^3} \\+ 15 \left( \frac{2}{3}\right)^2 \left( \frac{1}{3} \right)^4+6 \left( \frac{2}{3}\right)^1 \left( \frac{1}{3} \right)^5 + \left( \frac{1}{3}\right)^6 . \end{eqnarray*}

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .