You select a card at random from the cards that make up the word "replacement". Without replacing the card, you choose a second card. Find the probability of choosing a consonant and then an "e". There is 1 letter for each card.

Answer 1

To solve this problem, we need to determine the probability of first drawing a consonant and then drawing the letter "e" from the remaining cards. Let's break this down step-by-step.

The word "replacement" consists of the following 11 letters:
\[ r, e, p, l, a, c, e, m, e, n, t \]

First, we'll count the number of consonants and the number of "e"s in the word.

**Consonants:**
The consonants in "replacement" are:
\[ r, p, l, c, m, n, t \]

There are 7 consonants in total.

**The letter 'e':**
There are 3 "e"s in the word.

### Step-by-Step Calculation:

1. **Determine the probability of drawing a consonant first.**
The probability \( P(\text{Consonant}) \) is the number of consonants divided by the total number of letters:
\[
P(\text{Consonant}) = \frac{7}{11}
\]

2. **After drawing a consonant, we need to determine the probability of drawing an "e".**
Once we have drawn a consonant, there will be 10 cards left (since we don’t replace the card).

The number of "e"s remains the same because we haven't drawn an "e" yet, so there are still 3 "e"s remaining.

The probability \( P(\text{'e'}) \) after drawing a consonant is the number of "e"s divided by the remaining number of cards:
\[
P(\text{'e'}) = \frac{3}{10}
\]

3. **Combine the two probabilities.**
The combined probability of both events occurring in sequence (drawing a consonant first and then drawing an "e") is the product of the two individual probabilities:
\[
P(\text{Consonant and then 'e'}) = P(\text{Consonant}) \times P(\text{'e'})
\]
\[
P(\text{Consonant and then 'e'}) = \frac{7}{11} \times \frac{3}{10} = \frac{7 \times 3}{11 \times 10} = \frac{21}{110}
\]

Thus, the probability of choosing a consonant first and then an "e" from the remaining cards is \( \frac{21}{110} \).