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$\\displaystyle\\frac{\\var{red+green}}{\\var{total}}$

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$\\displaystyle\\frac{\\var{blue}}{\\var{green+red}}$

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

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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.

\n

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

\n

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

\n

\n

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.

\n

The total number of balls in the bag before the chosen ball is removed is 

\n

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

\n

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

\n

Therefore, the probability of the chosen ball being blue is

\n

\\[
P(\\text{blue}) = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}} = \\displaystyle\\frac{\\var{blue}}{\\var{total}}
\\]

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number of red balls in part c

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number of green balls in part c.

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total number of balls in part c

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number of blue balls in part c

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A bag contains balls of three different colours. You're told how many there are of each, and asked the probability of picking a ball of a particular colour.

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