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GMAT Statistics and Probability — practice questions — page 6

Explore 198 free Statistics and Probability questions in GMAT Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Statistics and Probability questions and exercises with answers

Showing questions 151–180 of 198.

For outcomes w1, ..., wn of S, the sum of P(wi) equalsRead the answer and worked solution →A coin has P(H) = 1/4. For a valid assignment, P(T) must beRead the answer and worked solution →A coin is assigned P(H) = 0.3 under the axioms. Then P(T) isRead the answer and worked solution →Six outcomes get probabilities 0.1, 0.2, 0.3, 0.4, 0.5 and 0.6. Their sum isRead the answer and worked solution →Giving an outcome a negative probability is invalid because probabilities must beRead the answer and worked solution →Three pens are tested, with 8 equally likely good/bad outcomes. P(exactly one bad) isRead the answer and worked solution →A card is drawn from a well shuffled 52-card deck. P(a diamond) isRead the answer and worked solution →If a card is a diamond with probability 1/4, P(not a diamond) isRead the answer and worked solution →If P(A) = 0.35, then the probability of 'not A' isRead the answer and worked solution →If P(A) = 0.4, P(B) = 0.5 and P(A and B) = 0.2, then P(A or B) isRead the answer and worked solution →For mutually exclusive A and B, P(A) = 0.3 and P(B) = 0.45. P(A or B) isRead the answer and worked solution →If P(A) = 0.5, P(B) = 0.6 and P(A or B) = 0.8, then P(A and B) isRead the answer and worked solution →In P(A) + P(B), the sample points of A and B together are countedRead the answer and worked solution →A fair die is rolled. P(a number less than 3) isRead the answer and worked solution →Two fair coins are tossed. P(exactly one head) isRead the answer and worked solution →Two fair dice are thrown. P(sum greater than 11) isRead the answer and worked solution →A card is drawn from ten cards numbered 1 to 10. P(2, 4, 6 or 8) isRead the answer and worked solution →Two dice are thrown. P(first die is 6 and sum is at least 11) isRead the answer and worked solution →A bag has 5 red, 3 green and 2 yellow marbles. P(red or yellow) isRead the answer and worked solution →If P(A) = 0.6 and P(B) = 0.7, the smallest possible P(A and B) isRead the answer and worked solution →A spinner lands on red in 30 of 120 trials. What is the experimental probability of red?Read the answer and worked solution →One ball is drawn uniformly from 9 red and 1 blue ball. Find P(blue).Read the answer and worked solution →A fair six-sided die shows 1 to 6. What is P(a score above 4)?Read the answer and worked solution →A card is drawn uniformly from a standard 52-card pack. What is P(a king)?Read the answer and worked solution →Pick an integer uniformly from 1 to 12. What is P(a factor of 12)?Read the answer and worked solution →A fair spinner has three equal sectors labelled A, A, B. Find P(A).Read the answer and worked solution →With all 52 standard cards equally likely, find the chance of drawing either black ace.Read the answer and worked solution →A month is chosen uniformly from the 12 months. What is P(it has exactly 30 days)?Read the answer and worked solution →A fair coin gave ten heads in a row. Its next toss is independent. Find P(head).Read the answer and worked solution →Find the chance that the product of two fair die scores (1 to 6) is odd.Read the answer and worked solution →

Worked practice questions

For outcomes w1, ..., wn of S, the sum of P(wi) equals

  1. 0
  2. 1
  3. 1/n
  4. n
Show answer and solution

Correct answer: B. 1

The outcomes together make up S, and P(S) is 1, so their probabilities sum to 1.

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A coin has P(H) = 1/4. For a valid assignment, P(T) must be

  1. 1/2
  2. 1/4
  3. 3/4
  4. 1
Show answer and solution

Correct answer: C. 3/4

P(H) + P(T) must equal 1, so P(T) = 1 - 1/4 = 3/4.

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A coin is assigned P(H) = 0.3 under the axioms. Then P(T) is

  1. 0.3
  2. 0.6
  3. 0.7
  4. 0.4
Show answer and solution

Correct answer: C. 0.7

With only two outcomes the probabilities sum to 1, so P(T) = 1 - 0.3 = 0.7.

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