Probability Basics and Listing Outcomes

Since $A$, $B$, and $C$ are the only possible outcomes, their probabilities must add up to $1$.

Therefore, if we take both $text{P(A)}$ and $text{P(B)}$ away from $100$ or $1$, we can work out $text{P(C)}$, the probability of the wheel stopping section $C$.

The only issue presented to us in this question is that the probability of the wheel stopping in section $A$ is given as a percentage, whereas for section $B$ it is expressed as a fraction. We will therefore need to convert either the fraction to the percentage or the percentage to the fraction so that the two probabilities are expressed in the same way.

Although you can turn a percentage to a fraction quite easily, in this question it is probably easier to convert the fraction $frac{2}{5}$ to a percentage (provided you know what the percentage value of $frac{1}{5}$ is):

$frac{1}{5} = 20$

so

$frac{2}{5} = 40$

Since the probability of the wheel stopping in section $A$ is $55$, and the probability of the wheel stopping in section $B$ is $40$, that means we can calculate the probability of the wheel stopping in either section $A$ or $B$:

$text{P(A)}+text{P(B)}= 55$

This therefore means that chance of the wheel stopping in section $C$ can be calculated as follows:

$100$

(An answer expressed as a decimal ($0.05$) or as a fraction $frac{1}{20}$ is also acceptable.)

Save your answers with

Gold Standard Education

Sign Up Now

a)  We know that the probability of Jimmy watching a romantic comedy is $0.56$.  Therefore we can easily calculate the probability of Jimmy not watching a romantic comedy:

$1 – 0.56 = 0.44$

This figure of $0.44$, the probability of Jimmy not watching a romantic comedy, is the same probability as Jimmy watching either a horror film or a sci-fi movie.

Since the probability of Jimmy watching a sci-fi movie or a horror film is equal, then the probability of Jimmy watching a sci-fi movie must be half of this amount:

$0.44 div2 = 0.22$

b)  From part a), we know that the probability of Jimmy watching a sci-fi movie is $0.22$.

The probability of Jimmy watching a romantic comedy is $0.56$.

In order to calculate the probability of Jimmy watching either a romantic comedy or a sci-fi movie we need to add the probabilities of each, since this is an either / or scenario:

$0.22 + 0.56 = 0.78$

Save your answers with

Gold Standard Education

Sign Up Now

If Macy chooses black trousers, then she has $3$ choices for the colour of her jumper, so the $3$ possible outcomes are:

BO,  BY,  BW

If she chooses navy trousers, then the $3$ possible outcomes are:

NO,  NY,  NW

Finally, if she chooses purple trousers, then the remaining possible outcomes are:

PO,  PY,  PW

BY was already given in the question, so the full list of other possible outcomes is:

BO,  BW,  NO,  NY, NW,  PO,  PY,  PW

Save your answers with

Gold Standard Education

Sign Up Now

a)  We know that the probability of selecting a red, blue or green bead must add up to $1$.  We know that the probability of selecting a red bead is $0.25$, so the probability of selecting either a blue or a green bead must be:

$1 – 0.25 = 0.75$

The problem in this question is that the probability of selecting a blue bead and the probability of selecting a green bead is not the same.  Selecting a blue bead has a probability $5x$ and selecting a green bead has a probability of $4x$.  This means that the probability of selecting a blue or green bead is:

 $5x + 4x = 9x$

Since we know that the probability of selecting a blue or green bead is $0.75$, we can therefore conclude that $9x = 0.75$

If

$9x = 0.75$

then

$x = 0.75 div 9 = dfrac{1}{12}$

If the probability of selecting a blue bead is $5x$, and $x = dfrac{1}{12}$, then the probability of selecting a blue bead is $dfrac{5}{12}$.

b) If the probability of selecting a green bead is $4x$, and $x = dfrac{1}{12}$, then the probability of selecting a blue bead is $dfrac{4}{12}$.  This fraction can then be simplified to $dfrac{1}{3}$

Save your answers with

Gold Standard Education

Sign Up Now

a)  We know that the tickets are only numbered between $1$ and $50$, so we need to work out how many of these fall into the category of being either a multiple of $5$ or an odd number.

There are $25$ odd numbers in total between $1$ and $50$:

$1, , 3, , 5, , 7, , 9, , 11,,  13,,  15,$ $17, , 19, , 21, , 23, , 25, , 27, , 29, , 31,$ $33, , 35, , 37, , 39, , 41, , 43, , 45, , 47, , 49$

There are $10$ multiples of $5$ between $1$ and $50$: $5, , 10, , 15, , 20, , 25, , 30, , 35, , 40, , 45, ,  50$

However, some of these multiples of $5$ also feature on the odd number list, so cannot be counted twice.  So, ignoring the odd multiples of $5$, there are only $5$ multiples of $5$ remaining.

$25 text{ odd numbers} + 5 text{ (even) multiples of } 5 = 30 text{ numbers in total}$

If $30$ of the $50$ numbers fall into the category of being either odd or a multiple of $5$, then as a fraction we can express this as:

$dfrac{30}{50}$ which can be simplified to $dfrac{3}{5}$

It is also perfectly acceptable to express the probability as a decimal or as a percentage:

$dfrac{3}{5}$ as a decimal is $3 div 5 = 0.6$

$dfrac{3}{5}$ as a percentage is $3 div 5 times 100 = 60$

b)  The factors of $48$ are as follows:

$1$ and $48$

$2$ and $24$

$3$ and $16$

$4$ and $12$

$6$ and $8$

(If you are ever working out the factors of a number, write them in pairs, and start at $1$.)

This means that $10$ numbers out of the $50$ in the hat are factors of $48$.  We can express this as a fraction:

$dfrac{10}{50}$ which can be simplified to $dfrac{1}{5}$

It is also perfectly acceptable to express the probability as a decimal or as a percentage:

$dfrac{1}{5}$ as a decimal is $1 div 5 = 0.2$

$dfrac{1}{5}$ as a percentage is $1 div 5 times 100 = 20$

Save your answers with

Gold Standard Education

Sign Up Now