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a)

For events where each outcome is equally likely to occur, we can calculate the probability using the formula

\\[\\text{Probability of an event} = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}.\\]

When we roll a fair $6$-sided die, we are equally likely to roll any of the numbers from $1$ to $6$ on the die as each number only appears once on the die.

i)

To find the probability of rolling an even number, we use the formula above.

The total number of outcomes is $6$ (as we are using a fair $6$-sided die).

Now we must think how many outcomes result in an even number being rolled.

There are $3$ possible outcomes where we roll an even number on the die:

we roll a $2$
we roll a $4$
we roll a $6$

Using the formula for probability for equally likely outcomes, this means that

\\[
\\begin{align}
P(\\text{rolling an even number}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\
&= \\displaystyle\\frac{3}{6}\\\\
&= \\displaystyle\\frac{1}{2}.
\\end{align}
\\]

ii)

To find the probability of not rolling a $\\var{die1}$ or a $\\var{die2}$, we use the same formula again.

The total number of outcomes is still $6$.

Here, we have $4$ possible outcomes where we don't roll a $\\var{die1}$ or a $\\var{die2}$ (i.e where we roll any of the other numbers on the die):

we roll a $\\var{die3}$
we roll a $\\var{die4}$
we roll a $\\var{die5}$
we roll a $\\var{die6}$

Using the formula,

\\[
\\begin{align}
P(\\text{not rolling a $\\var{die1}$ or a $\\var{die2}$}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\
&= \\displaystyle\\frac{4}{6}\\\\
&= \\displaystyle\\frac{2}{3}.
\\end{align}
\\]

b)

In this question, we are finding the probability of getting tails when an unbiased coin is flipped.

The probability of getting either heads or tails is equally likely, which means that we can use the same formula as we used throughout part a).

When we flip an unbiased coin there are $2$ possible outcomes: heads or tails.

There is only $1$ outcome where we obtain tails.

Using the formula,

\\[
\\begin{align}
P(\\text{tails}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\
&= \\displaystyle\\frac{1}{2}.
\\end{align}
\\]

c)

We are told that the bag contains $\\var{red}$ red balls, $\\var{blue}$ blue balls and $\\var{green}$ green balls and that one ball is removed from the bag at random. Hence, the total number of balls in the bag before the chosen ball is removed is

\\[\\var{red}+\\var{blue}+\\var{green} = \\var{total}.\\]

As the ball is being removed randomly from the bag, there is an equal probability of selecting any one of the $\\var{total}$ balls.

Therefore, using the same formula as in parts a) and b), we can calculate the probability of the chosen ball being blue as

\\[
\\begin{align}
P(\\text{blue}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\
&= \\displaystyle\\frac{\\var{blue}}{\\var{total}}.
\\end{align}
\\]

So, $P(\\text{blue}) = \\displaystyle\\simplify{{blue}/{total}}$.

", "statement": "

For equally likely outcomes, you can calculate the probability of a particular event occurring by using the formula

$\\text{Probability of an event} = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}$.

For example, rolling a fair $6$-sided die has $6$ possible outcomes which are given in the table.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Outcomes

We roll a $1$

We roll a $2$

We roll a $3$

We roll a $4$

We roll a $5$

We roll a $6$

Let's say we want to find the probability of rolling a $2$. From the list above, there is only one outcome which involves a $2$ being rolled, so the number of favourable outcomes is $1$.

When we roll a fair $6$-sided die, we are equally likely to roll any of the numbers from $1$ to $6$ on the die as each number only appears once on the die.

Hence using the above formula,

\\[
\\begin{align}
P(\\text{rolling a $2$}) &= \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}\\\\
&= \\displaystyle\\frac{1}{6}.
\\end{align}
\\]

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Potential number for a) ii)

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number of green balls in part c.

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Basic probability skills involving coins, dice and other events with equally likely outcomes.

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$1$

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$\\displaystyle\\frac{2}{3}$

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$\\displaystyle\\frac{1}{2}$

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$\\displaystyle\\frac{1}{3}$

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$\\displaystyle\\frac{2}{3}$

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$\\displaystyle\\frac{1}{3}$

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$\\displaystyle\\frac{1}{4}$

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$0$

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Choose the correct probability for each outcome when you roll a fair, $6$-sided die. Remember to reduce any fractions into their simplest form.

i) $P(\\text{rolling an even number})$

[[0]]

ii) $P(\\text{not rolling a $\\var{die1}$ or a $\\var{die2}$})$

[[1]]

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$1$

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$\\displaystyle\\frac{2}{3}$

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$\\displaystyle\\frac{1}{2}$

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$\\displaystyle\\frac{1}{3}$

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An unbiased coin is flipped. What is the probability of getting a tails?

[[0]]

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$\\displaystyle\\var[fractionnumbers]{({red}+{green})/{total}}$

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$\\displaystyle\\var[fractionnumbers]{{blue}/({green}+{red})}$

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$\\displaystyle\\var[fractionnumbers]{{blue}/{total}}$

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$\\displaystyle\\frac{1}{\\var{blue}}$

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$\\displaystyle\\frac{1}{\\var{total}}$

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A bag contains $\\var{red}$ red balls, $\\var{blue}$ blue balls and $\\var{green}$ green balls. One ball is removed from the bag at random. What is the probability that the chosen ball will be blue? Remember to reduce any fractions into their simplest form.

[[0]]

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