Question

I just need help cause if I don’t I don’t know what imma do

Answer 1

Answer:

$P(32) = \frac{1}{90}$

$P(Odd) = \frac{1}{2}$

$P(Multiples\ 5) = \frac{1}{5}$

Step-by-step explanation:

Given

Sample Space = 10 to 99

First, we calculate the sample size (n)

$n = 99 - 10 + 1$

$n = 90$

Solving (a): P(32)

In 10 to 99, there is only 1 32.

So,

$P(32) = \frac{n(32)}{n}$

$P(32) = \frac{1}{90}$

Solving (b): P(Odd)

There are 45 odd numbers between 10 and 99

i.e.

$n(Odd) = 45$

So:

$P(Odd) = \frac{n(Odd)}{n}$

$P(Odd) = \frac{45}{90}$

$P(Odd) = \frac{1}{2}$

Solving (c): P(Multiple of 5)

There are 18 multiples of 5 between 10 and 99

i.e

$P(Multiples\ 5) = \frac{18}{90}$

$P(Multiples\ 5) = \frac{1}{5}$