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At a highschool in a class of 235 students, 90 pass Mathematics, 85 pass English and 78 pass both courses. What’s the probability that a random student passed either Mathematics or English or both? ((Union question)

Sagot :

Answer:

  41.3%

Step-by-step explanation:

The probability of interest is found by dividing the number of students passing either course by the total number of students.

The number passing only one of the courses is the number passing that course, less the number passing both courses. This is shown in the attached 2-way table.

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The number passing either course will be the sum of the number passing one course (including the number passing both courses), and the number passing only the other course. It can also be found as the sum of the numbers passing either course, less the number passing both courses (since the latter would be counted twice).

Here those passing either (or both) Math or English will be ...

  90 +(85 -78) = 85 +(90 -78) = (90 +85) -78 = 97

So, the probability is ...

  p(passing either M or E) = 97/235 = 0.41277 ≈ 41.3%

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