Question

1. you select a card at random from the cards that make up the word

eplacement\. 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.

Explanation:

Step1: Count total letters and consonants

The word "replacement" has 11 letters. The consonants are r, p, l, c, m, n, t, so there are 8 consonants.

Step2: Calculate probability of first - draw (consonant)

The probability of choosing a consonant first is \(P(\text{consonant})=\frac{8}{11}\).

Step3: Calculate probability of second - draw (e)

After drawing a consonant, there are 10 letters left. There are 2 'e's in the word "replacement". So the probability of choosing an 'e' second is \(P(e)=\frac{2}{10}=\frac{1}{5}\).

Step4: Use multiplication rule for dependent events

The probability of both events (choosing a consonant then an 'e') is \(P = \frac{8}{11}\times\frac{1}{5}=\frac{8}{55}\). But wait, let's check again.
Wait, another approach:
The total number of ways to choose 2 letters without replacement from 11 letters is \(n(S)=A_{11}^2=\frac{11!}{(11 - 2)!}=11\times10 = 110\)
The number of ways to choose a consonant first (8 choices) and then an 'e' (2 choices) is \(n(A)=8\times2=16\)

Step5: Calculate the probability

The probability \(P=\frac{n(A)}{n(S)}=\frac{16}{110}=\frac{8}{55}\). Wait, no, wait the options:
Let's re - count the letters in "replacement":
The word "replacement" has letters: r, e, p, l, a, c, e, m, e, n, t. So there are 11 letters. Consonants: r, p, l, c, m, n, t (7 consonants). 'e's: 3.
The total number of ways to choose 2 non - replacement letters: \(n(S)=11\times10 = 110\)
The number of ways to choose a consonant first (7 choices) and then an 'e' (3 choices) is \(n(A)=7\times3 = 21\)
The probability \(P=\frac{21}{110}\)

Answer:

\(\frac{21}{110}\)