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Karen has 12 cards, each written with a letter from the word PENNSYLVANIA. She picks a card at random. Without replacing she picks another card at random. Find the probability she picks two vowels.

I was sure the answer is $\frac{5}{12}\times\frac{4}{11}$ :-)

Aaa Lol_dude
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Gamliel Beyderman
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    "Y" is not a vowel, even though it sometimes acts like one. – Daniel Mathias Jan 19 '22 at 23:50
  • Right idea, but see the above comment and make a quick fix – TheBestMagician Jan 19 '22 at 23:54
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    Here, I'd argue that $Y$ was a vowel, as the "syl" is its own syllable and syllables are meant to contain a vowel. – lulu Jan 19 '22 at 23:55
  • Gamliel did get $\frac{5}{12}$, so he probably took Y as a vowel. – Aaa Lol_dude Jan 20 '22 at 00:00
  • I concur with @lulu: as used in the English toponym of the Q, the $\mathtt{y}$ is definitely being used as a vowel and not a consonant, or at least moreso as a vowel than not-a-vowel, so ought to be treated as such, ergo validating the higher final probability calculation( of two distinct considered interpretations). – user946772 Jan 20 '22 at 03:03

2 Answers2

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With Y as vowel:

There are 5 vowels, picking a vowel first try gives you $\frac{5}{12}$ and then we have 11 letters and 4 vowels left, giving you $\frac{4}{11}$. Therefore, the answer would be $\frac{5}{12}\times\frac{4}{11}$, or $\frac{5}{33}$.

Without Y as vowel:

There are 4 vowels, picking a vowel first try gives you $\frac{4}{12}$, and then we will have 11 letters and 3 vowels left, giving you $\frac{3}{11}$. Therefore, the answer would be $\frac{4}{12}\times\frac{3}{11}$. or $\frac{1}{11}$.

Aaa Lol_dude
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  • $\underset{\tiny\text{As such, }}{∴ } \underset{\small\text{[the probability]}}{}=⟮(\underset{\small:[\text{ a vowel}]}{20} ⩒ \underset{\small :¬[\text{ a vowel}]}{12})/132\underset{\small\text{.. the final answer }\normalsize\text{( reduced)}}{⟯=⟮\underset{\small\text{iff  a token of syl}}{5/33} ⩒ \underset{\small\text{iff  not a token of syl}}{1/11}⟯}\underset{\small\text{, which is expressible equivalently as..}}{ ⇔ \underset{≘ 〈⟮ ⊏ \text{syl} ⟯,⟮ \not⊏ \text{syl} ⟯〉}{∈_{#=1}{\frac{5}{33},\frac{1}{11}}}}$. – user946772 Jan 20 '22 at 02:42
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There are four vowels, which are E, A, I, and A, so there's $\frac{5}{12}$ chance that Karen will pick a vowel at the first try. At the second try, there will be three vowels and eleven cards left, so we multiply $\frac{5}{12}$ by $\frac{3}{11}$ and you will get $\frac{1}{11}$

  • Your numbers are inconsistent. Since you are not taking Y as a vowel, you have four vowels, so your first probability should be $4/12$. – N. F. Taussig Jan 20 '22 at 00:15