What are the odds?

Posted May 15, 2013

by Lensyl Urbano

Three successful oregano plants out of four cuttings.

My middle school class stumbled upon a nice probability problem that I might just use in my pre-calculus exam.

Oregano

You see, four of my students took cuttings of oregano plants from the pre-school herb gardens when we were studying asexual reproduction, but only three of them grew roots.

That gives a 75% success rate.

P[success] = 0.75

The next time I do it, however, I want each student to have some success. If I have them take two cuttings that should increase the chances that at least plant will grow. And we can quantify this.

If the probability of success for each cutting is 75% then the probability of failure is 25%.

P[failure] = 0.25

Given a probability of one plant failing of 25%, the probability of both plants failing is the probability of one plant failing and the other plant failing, which, mathematically, is:

P[2 failures] = P[failure] x P[failure]

P[2 failures] = 0.25 x 0.25

P[2 failures] = 0.252

P[2 failures] = 0.0625

So the probability of at least one plant growing is the opposite of the probability of two failures:

P[at least 1 success of two plants] = 1 – P[2 failures]

P[at least 1 success of two plants] = 1 – 0.0625

P[at least 1 success of two plants] = 0.9375

which is about 94%.

If I had the students plant three cuttings instead, then the probability of at least one success would be:

P[at least 1 success of 3 plants] = 1 – P[3 failures]

P[at least 1 success of 3 plants] = 1 – 0.253

P[at least 1 success of 3 plants] = 1 – 0.02

P[at least 1 success of 3 plants] = 0.98

So 98%.

This means that if I had a class of 100 students then I would expect only two students to not have any cuttings grow roots.

An Equation

In fact, from the work I’ve done here, I can write an equation linking the number of cuttings (let’s call this n) and the probability of success (P). I did this with my pre-calculus class today:

P = 1 - 0.25^n

so if each student tries 4 cuttings, the probability of success is:

P = 1 – 0.254

P = 1 – 0.004

P = 0.996

Which is 96.6%, which would make me pretty confident that at least one will grow roots.

However, what if I knew the probability of success I needed and then wanted to back calculate the number of plants I’d need to achieve that probability, I could rewrite the equation solving for n:

Start with:
P = 1 - 0.25^n

isolate the term with n by moving it to the other side of the equation, and switching P as well:
0.25^n = 1 - P

take the natural log of both sides (we could take the log to the base 10 if we wanted, or any other base log, but ln should work just as well).

\ln{(0.25^n)} = \ln{(1 - P)}

now use the rules of logarithms to bring the exponent down into a multiplication:

n \ln{(0.25)} = \ln{(1 - P)}

solving for n gives:

n  = \frac{\ln{(1 - P)}}{\ln{(0.25)}}

Note that the probability has to be given as a fraction (between 0 and 1), so 90% is P = 0.9. A few of my students made that mistake.

Lavender

Roots on two of three plants.

At the same time my students were making oregano cuttings, six of them were making cuttings of lavender. Only one of the cuttings grew.

So now I need to find out how many lavender cuttings each student will have to make for me to be 90% sure that at least one of their cuttings will grow roots.

Students can do the math (either by doing the calculations for each step, or by writing and solving the exponential equation you can deduce from the description above), but the graph below shows the results:

Probabilities of successfully growing a lavender cutting.

Citing this post: Urbano, L., 2013. What are the odds?, Retrieved February 23rd, 2018, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Leave a Reply

You must be logged in to post a comment.

Creative Commons License
Montessori Muddle by Montessori Muddle is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License.