# P.o.t.D.- 301 problems

Sunday 30 September 2018

This is just a short entry to highlight the fact that I have now managed to compose more than **300 problems** in my **Problem of the Day** (P.o.t.D.) lists - one list for an **SL Problem of the Day**, and another list for an **HL Problem of the Day**.

I am trying to think up problems that will continue to be relevant for the new IB mathematics courses which will start being taught in August 2019 - **Analysis & Approaches** and **Applications & Interpretation**. The Applications & Interpretation course (which I refer to as the 'Applications' course) will have a greater emphasis on the integration of technology in the learning and 'doing' of mathematics. One fact that underlines this is that all of the exams for the Applications course (2 exams for SL, 3 exams for HL) will allow (require) a student to have a graphic display calculator (GDC).

So, for my 301st problem (problem #151 in the SL list) I decided to compose a problem that would feature effective use of a GDC; nothing complicated, just making use of one of the basic tools on any GDC - but one that many students do not fully utilize. Here is the problem.

#### Problem of the Day #151 (SL)

For all of the adults in the Netherlands, the ratio of the number of people who speak more than two languages to the number of people who speak two or fewer languages is *m* to 7 (or \(\frac{m}{7}\)), where \(m \in {\mathbb{Z}^ + }\). It is known that if eight Dutch adults are chosen at random, then the probability that at least one of them speaks more than two languages is approximately 0.9424. Find the value of *m*.

__Solution__:

There are eight independent ‘trials’ where only one of two results can be obtained – i.e. a person either speaks more than two languages or does __not__ speak more than two languages. If *X* represents the random variable for the “the number of adults who speak more than two languages”, then *X* has a **binomial distribution** (assuming each selection of a Dutch adult is an independent event).

The fraction of Dutch adults that speak two or more languages is \(\frac{m}{{m + 7}}\). Thus, the probability that a randomly selected adult speaks more than two languages is \(\frac{m}{{m + 7}}\), and the probability that a randomly selected adult does **not** speak more than two languages is \(1 - \frac{m}{{m + 7}} = \frac{{m + 7}}{{m + 7}} - \frac{m}{{m + 7}} = \frac{7}{{m + 7}}\).

It is given that \({\rm{P}}\left( {X \ge 1} \right) = 0.9424\).

Either there is at least one person that speaks more than two languages or there is no one that speaks more than two languages. Hence, \({\rm{P}}\left( {X \ge 1} \right) + {\rm{P}}\left( {X = 0} \right) = 1\;\;\;\; \Rightarrow \;\;\;\;{\rm{P}}\left( {X \ge 1} \right) = 1 - {\rm{P}}\left( {X = 0} \right)\).

Thus, \({\rm{P}}\left( {X \ge 1} \right) = 1 - {\rm{P}}\left( {X = 0} \right) = 0.9424\).

Need to solve for *m* in the equation \(1 - {\left( {\frac{7}{{m + 7}}} \right)^8} = 0.9424\)

This can be done with the equation solver on a GDC.

Therefore, \(m = 3\).

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