WEBVTT

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Hi,
I'm Fiona and this is Tamila 2021 paper 2

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and question three.

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We're given a scenario in which a student
is chosen at random from a class.

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We're told each student is equally likely
to be chosen from this class,

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and we are asked which of the following
conditions out of 1/2 and three is or are

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necessary for the probability that the
student wears glasses to equal 4 out of

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15.

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And we are given options A to H to choose
from when we get to that point.

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Let's start by looking at this fraction 4
fifteenths.

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It's in its lowest terms or its simplest
terms.

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Now,
this could represent a class of 15

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students, four of which wear glasses.

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But we could also have an equivalent
fraction.

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Let's say we could have a class of 30
students and eight of them could wear

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glasses.

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We could also have a class of 45 students
with 12 of them wearing glasses.

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And that's just three of of many possible
scenarios that this probability could

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represent.

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Now let's look through each statement and
what we want to see is whether each of

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these statements is a necessary condition
for the probability that the student

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wearing glasses equals 4 fifteenths.

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So statement one or condition one tells
us that exactly 11 students in the class

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wear glasses.

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Well,
if we had this scenario here and we had

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15 students in a class,
four of which wear glasses,

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then that would mean that there are 11
who do not wear glasses.

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But we know that there are other
scenarios represented by other equivalent

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fractions.

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So this condition is not necessary.

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The second condition is that the number
of students in the class is divisible by

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three.

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And if we look at the probability given
in its lowest terms,

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then the denominator is divisible by
three.

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And so that means that the number of
students in the class must be divisible

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by three.

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Because human beings we have to be
counted in integers.

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We cannot have half of a human or a
fraction of a human.

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And so this condition is necessary.

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The third condition tells us that the
class contains 30 students and eight of

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them wear glasses.

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So that is touching on this scenario here.

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But again,
we know that is only one out of many

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possibilities and so this statement or
this condition is not necessary for this

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to hold.

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So the answer is that only condition 2 is
necessary for the probability that the

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students wear glasses to equal 4 out of
15.

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So C is the answer to this one statement
2 only.

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Now let us take a moment to reflect on
this question and in particular to

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discuss the idea of what do we mean by
necessary in mathematics.

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And the way we write this is that if a
statement A is necessary for a statement

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B,
we write it like this with a backwards

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arrow, and that means that when B is true,
then A is guaranteed.

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Another way of saying this is that A must
hold in order for B to be true.

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The Tamua website,
UATUK website has a helpful booklet on

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logic and proof,
and that booklet uses a circle diagram

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which I think is helpful.

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So I'll show it to you here and we can
discuss that.

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So we have two circles,
circle A and circle B and if we are

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inside circle A then that means statement
A is true.

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If we are inside circle B,
that means that statement B is true.

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Now,
this diagram depicts that A is necessary

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for B,
which means that you have to be inside A

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in order to have any chance of being
inside circle B.

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Another way of thinking about this is
that if you're inside circle B,

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then you're guaranteed to also be inside
circle A.

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Let's think of a familiar example.

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So in order for a number to be a multiple
of four,

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it is necessary for that number to be
even.

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But we can also think about how not all
even numbers are multiples of four.

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And so this is the kinds of ways of
thinking when we are thinking about the

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idea of necessary in mathematics.