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A boy is typing 26 letters of the alphabet. Every letter shows up only once! What is the probability that in at least one of the two words will show up: "open", "abc"

Suppose $\ A = $ the word 'open' shows up and $\ B = $ the word 'abc' shows up.

$$\ P(A) = \frac{{23 \choose 1 } \cdot 22!}{26!} $$ there are 23 ways to arranging a sequence of 4 letters and thats why $\ {23 \choose 1} $ .

$$\ P(B) = \frac{{24 \choose 1}\cdot 23!}{26!} $$

$$\ P(A \cap B) = ?$$ I'm really not sure how to calculate probability of both events happening. and even trying to calculate the opposite ( the probability that both "open" and "abc" doesn't show).

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    $\begingroup$ Both words can't show up together : common letter! $\endgroup$ – Teresa Lisbon Nov 1 '18 at 10:47
  • $\begingroup$ Yes, sorry fixed. $\endgroup$ – bm1125 Nov 1 '18 at 10:47
  • $\begingroup$ Are you sure the original question had "bye", or is it "abc" there? $\endgroup$ – Parcly Taxel Nov 1 '18 at 10:50
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The probability that both events happen is zero, since the letter E appears in both words and there is no way to overlap the two words such that only one E appears.

Now suppose the words (which we assume are 4 and 3 letters long) share no common letter. Then the number of letter strings with both words can be computed by treating those words as units, so that there are 21 units to permute – 21 possibilities. $P(A\cap B)$ is then $\frac{21!}{26!}$.

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