Question

The question is in the pic! also, it wants it in simplified form

Answer 1

We will have the following:

a) P(at most 7, then February):

`P=((7)/(10))((1)/(12))\Rightarrow P=(7)/(120)`
`\Rightarrow P=0.058333333333\ldots\Rightarrow p\approx0.058`

So, the probability of getting at most 7 and then February is approximately 5.8%.

b) P(4, ten a month that starts with the letter J):

*First: We have that there are 3 months of the year that start with the letter J, so:

`p=((1)/(10))((3)/(12))\Rightarrow p=((1)/(10))((1)/(4))`
`\Rightarrow p=(1)/(40)\Rightarrow p=0.025`

So, the probability of getting 4 and then a month of the year that starts with J is 2.5%.

c) P(unshaded, then a month that starts with the letter A):

*We have that there are 4 out of 10 shaded segments.

*We also have that there are 2 months of the year that starts with A.

`p=((6)/(10))((2)/(12))\Rightarrow p=(1)/(10)`
`\Rightarrow p=0.1`

So, the probability of getting a value of the unshaded area and then a month of the year that starts with A is 10%.

d) P(multiple of 3, then a month with exactly 30 days):

*We have that there are 3 multiples of three in the spinner.

*We have that 4 months of the year have exactly 30 days.

`p=((3)/(10))((4)/(12))\Rightarrow p=(1)/(10)`
`\Rightarrow p=0.1`

So, the probability of choosing a multiple of 3 and then a month with exactly 30 days is 10%.

***Answers in order***

a):

`p=(7)/(120)`

b):

`p=(1)/(40)`

c):

`p=(1)/(10)`

d):

`p=(1)/(10)`