Four-letter word contains no two consecutive equal letters.

The approach for (a) should work, but you have forgotten to subtract cases where there are two different repeated letters. There are three patterns for this: AABB, ABAB, ABBA. For each there are $5\times 4$ choices for the letters, so this is an extra $60$ to subtract.

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The problems with the first attempt for (b) aren't quite as the other answer suggests. You do actually count AABB type possibilities, since case 1 counts cases where the first two are equal and different to the third, and the fourth can be anything (equal to the first or third or neither). Then case 2 counts cases where the middle two are equal, and the first and fourth are not the same as the middle two (but can be the same as each other). Case 3 seems to be the same as case 1 in reverse, in which case you have actually now counted AABB twice.

What's missing are the cases where there are more than two consecutive equal letters: $(5\times1\times 1\times 4)+(5\times 4\times1\times 1)+(5\times1\times1\times1)=45$.

Together with the fact that you have overcounted AABB-type ($5\times 1\times 4\times 1=20$) this should give the same answer as your other approach.

You did (a) completely wrong. The number of four-letter words with no repeats is $5×4×3×2=120$, so the probability is $\frac{120}{625}=\frac{24}{125}$. The mistake is that words with two repeats of two letters have not been accounted for.

For (b), the problem with the first approach is that it fails to consider cases where both the first two and last two letters are the same (e.g. AABB). The second approach gives the right answer.

some of your answers are incorrect, and some also use an unnecessarily complex approach.

(a) no. of words with no repeats is simply $P^5_4 = 5\cdot4\cdot3\cdot2 =120$

(b) no. of words with no two consecutive equal letters = $5\cdot4\cdot4\cdot4 = 320\quad$

You can then compute probabilities in each case by dividing by $5^4$