Question

Text & Tests 1 Junior Cycle Mathematics

3. Let U = {the natural number from 1 to 20}
T = {multiples of 3}
F = {multiples of 4}
(a) Draw a Venn diagram of this information.
(b) If a number is chosen at random from U, find:
(i) P(T) (ii) P(F) (ii) P(T") (iv) P(F') (V) PITNF)
(vi) P(TUF)
4. Let U = {months of the year}

Answer 1

a)

b) i) p(T) = 0.3

b) ii) p(F) = 0.25

b) iii) p(T') = 0.7

b) iv) p(F') = 0.75

b) v)
`p(T\cap F)=0.05`

b) vi)
`p(T\cup F)=0.5`

Explanation:

a)

The Venn diagram is shown in the figure.

In the Venn diagram, we have represented:

- The set of events U, which consists of all the natural numbers from 1 to 20

- The set of events T, which consists of all the numbers multiples of 3

- The set of events F, which consists of all the numbers multiples of 4

We notice that:

- The set T only contains the following numbers: 3,6,9,12,15,18, which are the multiples of 3

- The set F contains only the following numbers: 4,8,12,16,20, which are the multiples of 4

The number 12 is in common between the two sets T and F.

b)

i) After choosing a number from U, here we want to find

P(T)

which represents the probability that the extracted number is part of set T (so, that it is a multiple of 3).

This probability is given by:

`P(T)=(n(T))/(n(U))`

where

`n(T)` is the number of numbers in set T

`n(U)` is the number of numbers in set U

Here we have:

`n(T)=6` - the multiples of 3 between 1 and 20 are only: 3,6,9,12,15,18

`n(U)=20` (the numbers from 1 to 20 are 20)

So,

`p(T)=(6)/(20)=0.3`

ii)

After choosing a number from U, here we want to find

P(F)

which represents the probability that the extracted number is part of set F (so, that it is a multiple of 4).

This probability here is given by:

`P(F)=(n(F))/(n(U))`

where:

`n(F)` is the number of numbers in set F

`n(U)` is the number of numbers in set U

Here we have:

`n(F)=5` - the multiples of 4 between 1 and 20 are only: 4,8,12,16,20

`n(U)=20` (the numbers from 1 to 20 are 20)

Therefore,

`p(F)=(5)/(20)=0.25`

iii)

In this part, after a choosing a number from U, we want to find

p(T')

which is the propability of the set of events complementary to T. In other words, we want to find the probability that the extracted number is NOT part of set T.

The complementary probability of a certain set can be found using

`p(T')=1-p(T)`

where

`p(T)` is the probability of set T to occur

In this problem, as we calculated in part i), we have

p(T) = 0.3

Therefore, the probability of the complementary of T is

`p(T') = 1 - 0.3 = 0.7`

iv)

Similarly to part iii), In this part, after a choosing a number from U, we want to find

p(F')

this is the propability of the set of events complementary to F: so, we want to find the probability that the extracted number is NOT part of set F.

The complementary probability of this set can be found using

`p(F')=1-p(F)`

where:

`p(F)` is the probability of set F to occur

As we calculated in part ii), here we have

p(F) = 0.25

Therefore, the probability of F not to occur is

`p(F') = 1 - 0.25 = 0.75`

v)

In this part, we want to find

`P(T\cap F)`

which is the probability that after choosing a number from U, this number belongs to both sets T and F. In other words, the probability that the number is multiple of 3 and 4 at the same time.

This probability is given by:

`p(T\cap F) = (n(T\cap F))/(n(U))`

where

The numerator is the number of numbers between 1 and 20 being at the same time multiple of 3 and 4

The denominator is the number of numbers from 1 to 20

Here we have:

`n(T\cap F) = 1`, because only 1 number (12) is multiple of 3 and 4 at the same time between 1 and 20

`n(U)=20`

Therefore, this probability is

`p(T\cap F)=(1)/(20)=0.05`

vi)

In this part, we want to find

`p(T\cup F)`

which is the probability that after choosing a number from U, this number belongs either to set T or set F. In other words, the probability that the number is either a multiple of 3 or a multiple of 4.

This probability is given by:

`p(T\cup F)=(n(T\cup F))/(n(U))`

where

`n(T\cup F)` is the number of numbers which are either multiples of 3 or 4

`n(U)` is the number of numbers between 1 and 20

Here we have:

`n(T\cup F)=10`, since the numbers which are either multiple of 3 and 4 are: 3,4,6,8,9,12,15,16,18,20

n(U) = 20

Therefore,

`p(T\cup F)=(10)/(20)=0.5`