# Question in probability ( is in the image)?

## See below:

#### Explanation:

In all of these questions, we're working with the probability of the wheel landing on a certain space or a kind of space.

The probability of these events occurring can be found by dividing the number of ways we can meet the conditions by the number of ways the events can transpire:

$\text{number of ways conditions are met"/"number of ways events can turn out}$

i - On the spin, we get an even sector

There are 10 sectors and 5 of them are even ($2 , 4 , 6 , 8 , 10$). And so we get:

$P \left(\text{even}\right) = \frac{5}{10} = \frac{1}{2}$

ii - On the spin, we get a shaded sector

There are 4 shaded sectors, so we get:

$P \left(\text{shaded}\right) = \frac{4}{10} = \frac{2}{5}$

iii - On the spin, we either a shaded sector or an even one

We already know the probability of getting an even number is $\frac{5}{10}$ and that of getting a shaded number is $\frac{4}{10}$. But we can't simply add these together - there might be numbers that are both even and shaded and so we have to subtract those out (otherwise they get double counted). So we have:

$P \left(\text{even" nn "shaded")=P("even")+P("shaded")-P("even"uu"shaded}\right)$

$P \left(\text{even"uu"shaded}\right) = \frac{2}{10} = \frac{1}{5}$ - this is the numbers 2 and 4. And so we can say:

$P \left(\text{even" nn "shaded}\right) = \frac{5}{10} + \frac{4}{10} - \frac{2}{10} = \frac{7}{10}$

iv - On the spin, the number is odd and shaded.

We've already worked this out for even numbers and it turns out we have the same number of odd numbers that are also shaded (the 1 and 3). So we have:

$P \left(\text{odd"uu"shaded}\right) = \frac{2}{10} = \frac{1}{5}$