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The table shows the educational attainment of the population of a certain country, ages 2525 and over, expressed in nfitions Find the probability that a randomly selected person, aged 2525 or over, has completed four years of high school only or is male\newlineMale\newlineFemale\newlineTotal\newline\newlineYears of High School\newlineYears of College\newlineTotal\newline\newlineLess than 44\newline44 only\newlineSome (less than 44)\newline44 or more\newline\newline2727\newline2424\newline2121\newline8686\newline\newline252500\newline2424\newline252522\newline2121\newline252544\newline\newline252555\newline252566\newline252577\newline252588\newline252599\newline\newlineThe probability is \newline4400

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Q. The table shows the educational attainment of the population of a certain country, ages 2525 and over, expressed in nfitions Find the probability that a randomly selected person, aged 2525 or over, has completed four years of high school only or is male\newlineMale\newlineFemale\newlineTotal\newline\newlineYears of High School\newlineYears of College\newlineTotal\newline\newlineLess than 44\newline44 only\newlineSome (less than 44)\newline44 or more\newline\newline2727\newline2424\newline2121\newline8686\newline\newline252500\newline2424\newline252522\newline2121\newline252544\newline\newline252555\newline252566\newline252577\newline252588\newline252599\newline\newlineThe probability is \newline4400
  1. Rephrase Problem: First, let's rephrase the "What is the probability that a randomly selected person, aged 2525 or over, has completed four years of high school only or is male?"
  2. Find Totals: To solve this problem, we need to find the total number of people who have completed four years of high school only and the total number of males. Then we will add these numbers together, making sure not to double-count males who have also completed four years of high school only.
  3. Calculate Males: From the table, we can see that 2424 females and 2424 males have completed four years of high school only, making a total of 4848 people.
  4. Avoid Double-Counting: Next, we need to find the total number of males. Adding the males across all categories of educational attainment, we get 2727 (Less than 44 years of high school) + 2424 (44 years of high school only) + 2121 (Some college, less than 44 years) + 8686 (44 or more years of college) = 158158 males.
  5. Combine Numbers: Now, we need to subtract the number of males who have completed four years of high school only from the total number of males to avoid double-counting. Since there are 2424 males who have completed four years of high school only, we subtract this from the total number of males: 15824=134158 - 24 = 134 males who have not completed four years of high school only.
  6. Calculate Probability: We then add the number of people who have completed four years of high school only 4848 to the number of males who have not completed four years of high school only 134134 to get the total number of people who meet at least one of the conditions: 48+134=18248 + 134 = 182.
  7. Correct Total Population: Finally, we need to find the probability. The total population aged 2525 and over is the sum of all individuals in the table, which is 166166. The probability is the number of favorable outcomes (182182) divided by the total number of outcomes (166166).
  8. Recalculate Probability: Let's correct the total population by adding the total number of males and females: 158158 (total males) + 8080 (total females) = 238238.
  9. Simplify Fraction: Now we can calculate the correct probability. The probability is the number of favorable outcomes 182182 divided by the total number of outcomes 238238. This simplifies to the fraction 182238\frac{182}{238}.
  10. Simplify Fraction: Now we can calculate the correct probability. The probability is the number of favorable outcomes 182182 divided by the total number of outcomes 238238. This simplifies to the fraction 182238\frac{182}{238}.To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor. The greatest common divisor of 182182 and 238238 is 22. Dividing both by 22, we get 91119\frac{91}{119}.

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