A word game has 100 tiles, 98 of which are letters and 2 of which are blank. The numbers of tiles of each letter are shown. You randomly draw 1 tile, set it aside, and then randomly draw another tile.Event A: The first tile is a consonant.Event B: The second tile is a vowel.Find the probability that both events A and B will occur. Round your answer to the nearest tenth.about $\%$

To find the probability that both events A and B will occur, we need to follow these steps:

1. Determine the total number of consonants and vowels:

   • Let's assume the number of consonants is $C$ and the number of vowels is $V$.
   • The total number of tiles is 100, with 2 being blank, so there are 98 letter tiles.

2. Calculate the probability of drawing a consonant first (Event A):

   • The probability of drawing a consonant first is $\frac{C}{100}$.

3. Calculate the probability of drawing a vowel second (Event B) given that the first tile was a consonant:

   • After drawing a consonant, there are 99 tiles left.
   • The number of vowels remains $V$.
   • The probability of drawing a vowel second is $\frac{V}{99}$.

4. Calculate the combined probability of both events A and B:

   • The combined probability is the product of the individual probabilities: $P(A \text{ and } B) = P(A) \times P(B|A) = \left( \frac{C}{100} \right) \times \left( \frac{V}{99} \right)$

5. Substitute the values of $C$ and $V$ into the formula:

   • Let's assume $C = 56$ and $V = 42$ (these values are hypothetical and should be replaced with the actual counts from the game).

6. Calculate the probability: $P(A \text{ and } B) = \left( \frac{56}{100} \right) \times \left( \frac{42}{99} \right)$ $P(A \text{ and } B) = 0.56 \times 0.4242 \approx 0.2376$

7. Convert the probability to a percentage and round to the nearest tenth: $0.2376 \times 100 \approx 23.8\%$

Therefore, the probability that both events A and B will occur is approximately 23.8%.