Let S be the set of all strings of length 3 over the alphabet {a,b,c}. If you choose randomly an element x of S, the probability that there is a vowel in x is approximately.
A. 1/3
B. 1/2
C. 2/3
D. 1
My answer is A, but I got it wrong. The correct one is C, and I don't know why?
Thank You
Consider drawing with replacement,
\begin{align} Pr(a \text{ appears in } x) &= 1 - Pr(a \text{ does not appear in }x) \\ &=1 - \left( \frac23\right)^3\\ &= 1-\frac8{27} \\ &= \frac{19}{27} \approx \frac{18}{27}=\frac23 \end{align}
Each of the three letters in $x$ is selected independently (with replacement) from an alphabet of $3$ options, one of which is a vowel and two of which are consonants. That is a total of $3^3$ equally probable outcomes, with $2^3$ of them containing only consonants, and (how many) do not.
The probability that there is at least one vowel in $x$ is: $\dfrac{?}{3^3}$
Donald Splutterwit has given a good hint in his comment that will lead you to the answer. Let me attempt to tell you why your answer (1/3) is wrong.
Your answer is correct for a single letter. Assume you have heavy coin which has probabilty 1/3 of coming with head when tossed. (That is a single toss) But when you toss is three times the probability of getting head at least once should increase. Same way a vowel (it is the letter a here) has probability of 1/3 when selected from $\{a,b,c\}$. But in a string a 3 letters the chances of a being present should shoot up. Theactual numerical calculations are given by others. I hope this intuitive explanation helps you understand.