Probability of 1 Letter E in 4 Randomly Selected Letters from 'ENCYCLOPAEDIA

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Homework Statement

Four letters are randomly selected from the word ENCYCLOPAEDIA. Find the probability that one letter E will occur in the selection of 4 letters.

Relevant Equations

[SUP]2[/SUP]C[SUB]1[/SUB] x [SUP]11[/SUP]C[SUB]3[/SUB]/[SUP]13[/SUP]C[SUB]4[/SUB] = 0.46

Four letters are randomly selected from the word ENCYCLOPAEDIA. Find the probability that one letter E will occur in the selection of 4 letters.

My calculation was

²C₁ x ¹¹C₃/¹³C₄ = 0.46

However, the answer is 2/33. I think the repeated letters might need to be accounted for. I divided the E term in the numerator by 2. I also divided the term for the remaining 11 letters by 4. I divided the denominator by 8 (2!x2!x2!), but this just canceled out the 2x4 in the numerator.
Thanks

 

[PeroK]

I agree with your answer. I did it using direct probabilies. First, the probability of getting an $E$ as the first letter, followed by three letters that are not $E$ is:
$$P(EXXX) = \frac 2{13} \cdot \frac{11}{12} \cdot \frac{10}{11} \cdot \frac{9}{10} = \frac 3 {26}$$
Then we have the same probability for $XEXX$ etc. So, the total probability of precisely one $E$ is $\frac{12}{26} = 0.46$.

Your method worked because you effectively labelled the duplicate letters $E_1, E_2$ and looked for either $E_1$ or $E_2$. And the same with the other duplicate letters.

So, we have three ways to calculate and, happily, they all give the same answer.

PS you can see that $2/33$ can't be right, because you have only three options: no E's, one E ort two E's. The last of these looks the least likely, so you'd expect quite a high probability of either no E's or one E.

You could, as an exercise, calculate those probabilities as well: no E's and two E's. And check it all adds up to 1.