# Suppose you throw a dart at the board shown in the diagram. Also suppose the dart was guaranteed to hit the board but your aim is such that you are unable to target a particular area. See below for rest of question ?

## What is the probability of your dart landing in the blue area?

Sep 4, 2016

#### Answer:

$\frac{7}{8}$

#### Explanation:

Looking at the wanted areas, we can see that the only areas which are not included are the 2 cream triangles, which together will form a rectangle measuring 3x5.

While we could do calculations and find all the various areas - looking at the sides of the trapezoids it can be seen that there can be 8 rectangles drawn inside the big rectangle.

Of these, only 1 rectangle is not part of the desirable area.

$P \left(\text{trap or blue}\right) = \frac{7}{8}$

Sep 6, 2016

#### Answer:

$\frac{7}{8}$

#### Explanation:

$\textcolor{b l u e}{\text{Important note}}$

Probability is another way of expressing counting.

Count every likely hood and you have all the likely hoods. As this is every thing it is considered as the whole of all of the events. Thus it is allocated the value of 1. The value 1 represents absolute certainty in that something from this is definitely going to occur. So if you split up the different events within the whole they can be represented as a fraction of the whole.

Thus probability is shown as fractional likely hoods. Add up all the fractions and you get 1

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$\textcolor{b l u e}{\text{Solving the question}}$

Blue or Trapezoid $\textcolor{red}{\text{excludes}}$ the two yellow triangles at bottom left and bottom right. Find the probability of hitting those and subtract that from 1 and you have the probability you need.

Using area to represent likely hood

The area of the whole is $10 \times 12 = 120$

The area of the 'bits' not included is $3 \times 5 = 15$

So the likely hood of hitting the area not interested in is $\frac{15}{120}$

So the likely hood of hitting the area we are interested in is:

$1 - \frac{15}{120} = \frac{7}{8}$