We will have the following:
a) P(at most 7, then February):
[tex]P=(\frac{7}{10})(\frac{1}{12})\Rightarrow P=\frac{7}{120}[/tex][tex]\Rightarrow P=0.058333333333\ldots\Rightarrow p\approx0.058[/tex]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:
[tex]p=(\frac{1}{10})(\frac{3}{12})\Rightarrow p=(\frac{1}{10})(\frac{1}{4})[/tex][tex]\Rightarrow p=\frac{1}{40}\Rightarrow p=0.025[/tex]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.
[tex]p=(\frac{6}{10})(\frac{2}{12})\Rightarrow p=\frac{1}{10}[/tex][tex]\Rightarrow p=0.1[/tex]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.
[tex]p=(\frac{3}{10})(\frac{4}{12})\Rightarrow p=\frac{1}{10}[/tex][tex]\Rightarrow p=0.1[/tex]So, the probability of choosing a multiple of 3 and then a month with exactly 30 days is 10%.
***Answers in order***
a):
[tex]p=\frac{7}{120}[/tex]b):
[tex]p=\frac{1}{40}[/tex]c):
[tex]p=\frac{1}{10}[/tex]d):
[tex]p=\frac{1}{10}[/tex]