Probability: Harder Probability - Using Combinations
Four letter tiles spelling the word "MATH" are picked up, shuffled, and set down in a random order forming a four-letter "word." What is the probability that the position of the "A" tile remains unchanged from the original word "MATH"?
Correct.
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There are four equally likely positions where the "A" tile can end up. In one of those positions, it remains unchanged. Thus the probability that the "A" tile ends up in the second position is $$\frac{1}{4}$$.
While it is longer, you can calculate the answer using combinatorics. Your denominator is the total number of ways in which the word "MATH" can be rearranged. It has four different letters. Thus, the number of arrangements is
>$$\text{Arrangements} = 4! = 4\times 3\times 2\times 1 = 24$$.
(The number of arrangements of $$n$$ different items is $$n!$$).
Now, if the position of the "A" tile does not change, then the first, third, and fourth positions are reserved for consonants, and the vowel A remains in the second position. The consonants "M," "T," and "H" can be rearranged in the first, third, and fourth positions in $$3! = 6$$ ways.
Therefore, the probability that the "A" tile doesn't change position is
> $$\text{Probability} = \frac{3!}{4!}=\frac{6}{24}=\frac{1}{4}$$.
Incorrect.
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Incorrect.
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Incorrect.
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Incorrect.
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$$\frac{1}{3}$$
$$\frac{1}{4}$$
$$\frac{1}{6}$$
$$\frac{1}{12}$$
$$\frac{1}{24}$$
