63. a) The word "ремонт" is made up of letters from a cut-out alphabet. Then the cards with individual letters are thoroughly shuffled, after which four of them are randomly drawn and laid out in a row in the order of drawing. What is the probability of getting the word "море" in this way?

b) A similar operation is performed with the letters that initially formed the word "папаха". What is the probability of getting the word "папа" in this way?
63. a) The considered experiment consists in sequentially drawing some four cards from a group of six cards. In this case, the first draw can result in six different outcomes (the first can be any of the six cards), the second in five different outcomes (the second can be any of the five cards remaining after the first draw), the third in four different outcomes, and the fourth in three different outcomes. By combining the six possible outcomes of the first draw with the five possible outcomes of the second draw, we get $6 \cdot 5 = 30$ possible outcomes for the sequence of two draws; similarly, for the sequence of three draws, we get $4 \cdot 30 = 120$ possible outcomes, and for the sequence of four draws, $3 \cdot 120 = 360$ possible outcomes. Thus, our experiment can result in 360 different outcomes; all these outcomes are equally probable.

The favorable outcome for us is the single outcome where the card with the letter $M$ is drawn first, the card with the letter $O$ is drawn second, the card with the letter $P$ is drawn third, and finally, the card with the letter $E$ is drawn fourth. Therefore, the desired probability is $\frac{1}{360} \approx 0.003$.

b) Here, the considered experiment again consists in sequentially drawing four cards from a group of six identical cards and, consequently, can have 360 equally probable outcomes. However, counting the number of favorable outcomes here is more complicated than in the case of problem a) due to the presence of identical letters in the word "papaha". Favorable outcomes here are those in which the first card drawn is any of the two cards with the letter p, the second is any of the three cards with the letter a, the third is the second card with the letter p (remember that one such card has already been drawn first), and the fourth is any of the two cards with the letter a remaining after the second draw. By combining the two possible outcomes of the first draw with the three possible outcomes of the second, one possible outcome of the third, and two possible outcomes of the fourth, we get a total of $2 \cdot 3 \cdot 1 \cdot 2 = 12$ different favorable outcomes. Therefore, the desired probability here is $\frac{12}{360} = \frac{1}{30} \approx 0.033$.