GMAT Book 3: Probability

Discrete probability, Independent events, Mutually exclusive events

Probability questions tested in the GMAT quant section include concepts such as independent events, mutually exclusive events, collectively exhaustive events, complementary events, and questions covering tossing coins, and rolling dice.

Syllabus Covered in Wizako's GMAT Book | Discrete Probability

The concepts in this topic are essentially an extension of the fundamentals learnt in Permutation Combination. Many of the probability questions are restricted permutation questions. Wizako's GMAT Math Lesson Book in this chapter covers the following concepts:

  1. Introduction to the concept of probability.
  2. Meaning of sample space and events along with illustrative examples to explain the same.
  3. Types of experiments in probability.
  4. Introduction to independent events, mutually exclusive events and probability of complimentary and exhaustive events.
  5. Explanation of compound events.
  6. Method to evaluate the probability of an event with illustrative example and shortcut methods.
  7. Geometric probability.
  8. 6 illustrative examples. 27 solved examples covering typical questions in tossing of coins, rolling of dice, picking cards from a pack of cards.
  9. 13 exercise problems with the answer key and also explanatory answers.
  10. An objective type test with 45 GMAT level multiple choice questions in the work book. An answer key and explanatory answer for all questions is provided.

Here is a typical solved example in Wizako's GMAT Book from this chapter

Sample Question

A bag contains 5 yellow balls and 6 orange balls. When 4 balls are drawn at random simultaneously from the bag, what is the probability that not all of the balls drawn are orange?

Explanatory Answer

Sample Space (Denominator) : Four balls can be drawn from a bag containing 11 balls in 11C4 ways

Event (Numerator) : The number of ways in which all four balls drawn will all be orange = 6C4.

Probability: The probability that all four balls drawn are orange
= \\frac{{^{6}}{C}_{4}}{{^{11}}{C}_{4}}) = \\frac{30}{330}) = \\frac{1}{11})

Therefore, the probability that not all of the balls drawn are orange = 1 - probability that all four are orange
= 1 - \\frac{1}{11})
= \\frac{10}{11})


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