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The probability that a teacher will give an unannounced test during any class is $\dfrac15$. If a student is absent twice then probability that he/she misses at least one test is

$\\ \hspace{5cm}$ a) $\dfrac23\ \quad $ b) $\dfrac45\ \quad$c) $\dfrac7{25}\ \quad $d) $\dfrac9{25}\ $

My attempt:

Probability of attending first test & missing $2$nd test $=\dfrac45\times\dfrac15=\dfrac4{25}$

Probability of missing first test & attending $2$nd test $=\dfrac15\times\dfrac45=\dfrac4{25}$

Probability of missing both the tests $=\dfrac15\times\dfrac15=\dfrac1{25}$

Total probability of missing at least one test $=\dfrac4{25}+\dfrac4{25}+\dfrac1{25}=\dfrac9{25}$

Can somebody please help me if I am wrong? Thanks.

Community
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jeanne clement
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    Yes, you are correct. Notice, a simple method is $$P_{\text{ Missing atleast one test }} = 1- P_{\text{ Missing no test }} = 1-(\frac{4}{5})^2=\frac{9}{25}$$ –  Dec 10 '17 at 15:42

2 Answers2

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Questions with "at least" are often great candidates for using the complement. That is: $$P_{\text{ Event A occurs }} = 1 - P_{\text{ Event A does not occur }}$$

So, $P_{\text{ miss at least 1 test }} = 1 - P_{\text{ miss no tests }}$ giving us:

$$\begin{align*} P_{\text{ miss no tests }} & = P_{\text{ no test on day 1 AND no test on day 2 }}\\ & = P_{\text{ no test on day 1 }} \times P_{\text{ no test on day 2 }}\\ & = \frac{4}{5} \times \frac{4}{5}\\ & = \frac{16}{25}\end{align*}$$

So, $$P_{\text{ miss at least 1 test }} = 1 - P_{\text{ miss no tests }} = 1 - \frac{16}{25} = \frac{9}{25}$$

N. F. Taussig
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Simon
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Another way to see this is that the probability of the student missing at least one test is the probability of missing test 1 + the probability of missing test 2 - the probability of them missing both tests.

The student has $\frac15 $chance of missing the first test, and $\frac15 $ chance of missing the second test. If you add them up you're left with $\frac 25$. HOWEVER there is a case that is being counted twice: It could happen that the student misses test 1 AND test 2. That case is being counted both when the student misses test 1 and when the student misses test 2. So now we have to substract this case once

The chance of a test happening on two consecutive days is $(\frac 15)^2 = \frac {1} {25}$.

So the answer is $\frac 25 - \frac {1} {25} = \frac {9} {25}$

The general case for this is $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Francisco José Letterio
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