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A bag contains 30 lettered tiles. There are 4 different letters in the bag. A tile is pulled from the bag and replaced 100 times. The results are:

The letter W is pulled 29 times.
The letter X is pulled 40 times.
The letter Y is pulled 11 times.
The letter Z is pulled 20 times.
Question
Based on this information, how many of each lettered tile are ,begin emphasis,most,end emphasis, likely in the bag? Enter the answers in the boxes.

Response area with 4 text input boxes
Letter Number of Tiles
W

X

Y

Z

1 answer

To determine the number of each lettered tile in the bag, we can use the results from the experiment to set up a proportion.

We have the following pulls from the bag:

  • W: 29 times
  • X: 40 times
  • Y: 11 times
  • Z: 20 times

The total number of pulls is:

\[ 29 + 40 + 11 + 20 = 100 \]

Now, we can find the experimental probability for each letter:

  • Probability of pulling W: \( \frac{29}{100} \)
  • Probability of pulling X: \( \frac{40}{100} \)
  • Probability of pulling Y: \( \frac{11}{100} \)
  • Probability of pulling Z: \( \frac{20}{100} \)

Next, we assume the number of tiles of each letter corresponds to these observed probabilities. Let \( n_W \), \( n_X \), \( n_Y \), and \( n_Z \) be the number of tiles for letters W, X, Y, and Z respectively. The total number of tiles is 30, thus:

\[ n_W + n_X + n_Y + n_Z = 30 \]

We can express the number of tiles in terms of probabilities:

  • \( n_W = 30 \cdot \frac{29}{100} = 8.7 \) (approximately 9)
  • \( n_X = 30 \cdot \frac{40}{100} = 12 \)
  • \( n_Y = 30 \cdot \frac{11}{100} = 3.3 \) (approximately 3)
  • \( n_Z = 30 \cdot \frac{20}{100} = 6 \)

Since the number of tiles must be whole numbers, we round \( n_W \) and \( n_Y \) to the nearest whole number, while ensuring that the total remains 30. Adjustments will need to be made because rounding can lead to exceeding or falling short of 30.

After rounding, we can check the distribution:

  • Let’s say we round \( n_W \) to 9 and \( n_Y \) to 3, which gives:
    • \( n_W = 9 \)
    • \( n_X = 12 \)
    • \( n_Y = 3 \)
    • \( n_Z = 6 \)

Now, let's verify the total: \[ 9 + 12 + 3 + 6 = 30 \]

This distribution works. Therefore, we can enter the following numbers:

Letter Number of Tiles

  • W: 9
  • X: 12
  • Y: 3
  • Z: 6