GRE® Probability Practice

Concept: Maximize and minimize the probability of an event.

This sample GRE quant practice question is a probability problem solving question. You have to compute the maximum and minimum probability of an event.

Question: Ms Li works at an office where the work timing is from 9:00 AM to 6:00 PM. 25% of a year she goes late to office and 35% of a year she leaves early from office. If P is the probability that she works at office the entire day then

  1. 0.25 ≤ P ≤ 0.35
  2. 0.25 ≤ P ≤ 0.65
  3. 0.4 ≤ P ≤ 0.65
  4. 0.35 ≤ P ≤ 0.4
  5. 0.1 ≤ P ≤ 0.6

Video Explanation

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Explanatory Answer

Use these hints to get the answer

  1. Consider the scenario that the days she reached office late were the days she left early to compute maximum probability of her having worked the entire day.
  2. Consider the scenario that all the days she reached office late were different from the days she left office early to compute the minimum probability of her having worked the entire day.

Scenario 1: Maximum probability of her having worked the entire day

The probability P that she worked the entire day will be maximized only when the number of days that she did not work the entire day is minimized.

If all the days that she reached office late coincided with the days she left office early, then the larger of the two sets will be the number of days she did not work the entire day.

i.e., if the 25% of the days when she reached office late falls under the 35% of days when she left office early, she would not have worked for the entire day for only 35% of the days.

In other words, if the days she reached office late is a subset of the days she left office early, P(A ∪ B) will be minimized.

Hence, for a minimum of 35% of the days she did not work the entire day.

Therefore, for a maximum 65% of the days she would have worked the entire day.

Scenario 2: Minimum probability of her having worked the entire day

The probability P that she worked the entire day will be minimized only when the number of days that she did not work the entire day is maximized.

If the days that she reached office late were different from the days she left office early, then the number of days she did not work the entire day will be maximized.

i.e., if the 25% of the days when she reached office late are different from the 35% of days when she left office early, she would not have worked the entire day for 25% + 35% = 60% of the days.

In other words, if the set of days she reached office late and the set of the days she left office early are disjoint sets, P(A ∪ B) will be maximized.

Hence, for a maximum of 60% of the days she did not work the entire day.

Therefore, for a minimum of 40% of the days she could have worked the entire day.

So, the range of values that P takes is 0.4 ≤ P ≤ 0.65

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