Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The table shows the educational attainment of a population, expressed in millions. Find the odds in favor and the odis against a randomly selected member of the population with four years (or more) of college








Less than 4


Years High


School







4 Years


Only







Some


Onden


than 4 years)







4 Years


College (or


More)




Total


Male
11
26
20
23
80


Female
16
28
22
21
87


Total
27
54
42
44
167




The odds in most reduced form, in favor of selecting a member of the population with four years (or more) of college are
(Simplity your answers)

The table shows the educational attainment of a population, expressed in millions. Find the odds in favor and the odis against a randomly selected member of the population with four years (or more) of college\newline\begin{tabular}{|l|c|c|c|c|c|}\newline\hline & \begin{tabular}{c} \newlineLess than 44 \\\newlineYears High \\\newlineSchool\newline\end{tabular} & \begin{tabular}{c} \newline44 Years \\\newlineOnly\newline\end{tabular} & \begin{tabular}{c} \newlineSome \\\newlineOnden \\\newlinethan 44 years)\newline\end{tabular} & \begin{tabular}{c}\newline44 Years \\\newlineCollege (or \\\newlineMore)\newline\end{tabular} & Total \\\newline\hline Male & 1111 & 2626 & 2020 & 2323 & 8080 \\\newline\hline Female & 1616 & 2828 & 2222 & 2121 & 8787 \\\newline\hline Total & 2727 & 5454 & 4242 & 4444 & 167167 \\\newline\hline\newline\end{tabular}\newlineThe odds in most reduced form, in favor of selecting a member of the population with four years (or more) of college are\newline(Simplity your answers)

Full solution

Q. The table shows the educational attainment of a population, expressed in millions. Find the odds in favor and the odis against a randomly selected member of the population with four years (or more) of college\newline\begin{tabular}{|l|c|c|c|c|c|}\newline\hline & \begin{tabular}{c} \newlineLess than 44 \\\newlineYears High \\\newlineSchool\newline\end{tabular} & \begin{tabular}{c} \newline44 Years \\\newlineOnly\newline\end{tabular} & \begin{tabular}{c} \newlineSome \\\newlineOnden \\\newlinethan 44 years)\newline\end{tabular} & \begin{tabular}{c}\newline44 Years \\\newlineCollege (or \\\newlineMore)\newline\end{tabular} & Total \\\newline\hline Male & 1111 & 2626 & 2020 & 2323 & 8080 \\\newline\hline Female & 1616 & 2828 & 2222 & 2121 & 8787 \\\newline\hline Total & 2727 & 5454 & 4242 & 4444 & 167167 \\\newline\hline\newline\end{tabular}\newlineThe odds in most reduced form, in favor of selecting a member of the population with four years (or more) of college are\newline(Simplity your answers)
  1. Determine College Education Count: First, we need to determine the number of people with four years (or more) of college. According to the table, there are 2323 million males and 2121 million females with this level of education, for a total of 4444 million people.
  2. Calculate Total Population: Next, we calculate the total population. The table shows that there are 167167 million people in total.
  3. Find Odds in Favor: Now, we find the odds in favor of selecting a member with four years (or more) of college. Odds in favor are calculated as the ratio of the number of favorable outcomes to the number of unfavorable outcomes. The favorable outcomes are the 4444 million with four years (or more) of college, and the unfavorable outcomes are the total population minus the favorable outcomes, which is 167167 million - 4444 million == 123123 million.
  4. Simplify Odds in Favor: The odds in favor are therefore 4444 to 123123. To simplify, we look for the greatest common divisor (GCD) of 4444 and 123123. The GCD of 4444 and 123123 is 11, so the odds in favor cannot be simplified further.
  5. Calculate Odds Against: Now, we calculate the odds against selecting a member with four years (or more) of college. Odds against are the inverse of odds in favor, which means we take the number of unfavorable outcomes over the number of favorable outcomes. The odds against are therefore 123123 to 4444.

More problems from Probability of independent and dependent events