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Question
the table shows the educational attainment of a population, expressed in millions. find the odds in favor and the odds against a randomly selected member of the population with four years (or more) of college.
| male | female | total | |
|---|---|---|---|
| 4 years high school only | 30 | 25 | 55 |
| some college (less than 4 years) | 25 | 18 | 43 |
| 4 years college (or more) | 25 | 21 | 46 |
| total | 93 | 78 | 171 |
the odds, in most reduced form, in favor of selecting a member of the population with four years (or more) of college are \\(\square:\square\\).
(simplify your answers.)
Step1: Calculate the number of people with four years (or more) of college
From the table, the total number of people with four years (or more) of college (the "More" row) is \(46\) million.
Step2: Calculate the number of people without four years (or more) of college
The total population is \(171\) million. The number of people without four years (or more) of college is \(171 - 46=125\) million.
Step3: Calculate the odds in favor
The odds in favor of an event \(E\) (selecting a person with four years (or more) of college) is given by \(\frac{n(E)}{n(\text{not }E)}\), where \(n(E)\) is the number of favorable outcomes and \(n(\text{not }E)\) is the number of unfavorable outcomes. So, the odds in favor is \(\frac{46}{125}\).
Step4: Calculate the odds against
The odds against an event \(E\) is \(\frac{n(\text{not }E)}{n(E)}\), so the odds against is \(\frac{125}{46}\).
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The odds in favor are \(\frac{46}{125}\) and the odds against are \(\frac{125}{46}\)