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A jar contains 5 green, 7 yellow, and 8 blue marbles. If one marble is drawn at

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Question: A jar contains 5 green, 7 yellow, and 8 blue marbles. If one marble is drawn at random, what is the probability that it is either green or yellow?

Options:

  1. 1/3
  2. 5/20
  3. 12/20
  4. 1/2

Correct Answer: 12/20

Solution: 

Total marbles = 5 + 7 + 8 = 20. Green or yellow = 5 + 7 = 12. Probability = 12/20 = 3/5.

A jar contains 5 green, 7 yellow, and 8 blue marbles. If one marble is drawn at

Practice Questions

Q1
A jar contains 5 green, 7 yellow, and 8 blue marbles. If one marble is drawn at random, what is the probability that it is either green or yellow?
  1. 1/3
  2. 5/20
  3. 12/20
  4. 1/2

Questions & Step-by-Step Solutions

A jar contains 5 green, 7 yellow, and 8 blue marbles. If one marble is drawn at random, what is the probability that it is either green or yellow?
  • Step 1: Count the number of green marbles in the jar. There are 5 green marbles.
  • Step 2: Count the number of yellow marbles in the jar. There are 7 yellow marbles.
  • Step 3: Count the number of blue marbles in the jar. There are 8 blue marbles.
  • Step 4: Add the total number of marbles together: 5 (green) + 7 (yellow) + 8 (blue) = 20 total marbles.
  • Step 5: Add the number of green and yellow marbles together: 5 (green) + 7 (yellow) = 12 marbles that are either green or yellow.
  • Step 6: To find the probability of drawing a green or yellow marble, divide the number of green or yellow marbles (12) by the total number of marbles (20): 12/20.
  • Step 7: Simplify the fraction 12/20 to its lowest terms: 12/20 = 3/5.
  • Probability – The likelihood of an event occurring, calculated as the number of favorable outcomes divided by the total number of possible outcomes.
  • Counting Outcomes – Understanding how to count the total number of items and the number of items that meet certain criteria.
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