Probability of the second player in selection of cards

Question

J and K decided to play a game . ten cards are marked with the letters of the word "RAFFLESIAN" respectively

these cards are placed into a bag. In the game, each player takes turn to draw 2 cards without replacement from the bag containing the ten cards. J stars first . the player who first draws 2 cards containing vowels wins.

1. calculate probability of J wining on his first draw
2. calculate probability of K wining on his first draw

Answer 1

I am presuming each player draws two cards at a time.

For J to draw two vowels, this means (4/10)*(3/9) = 2/15.

J can lose on the first try by drawing two consonants, a vowel followed by a consonant, or a consonant followed by a vowel.

For K to win, K must now draw 2 vowels.

J must not win. This can happen if J:

1. Draws 2 consonants:

(6/10)*(5/9) = 1/3

For K to win in this case, (1/3)*(4/8)*(3/7) = 1/14.

2. Draws a consonant, then a vowel:

(6/10)*(4/9) = 4/15.

For K to win in this case, (4/15)*(3/8)*(2/7) = 1/35.

3. Draws a vowel, then a consonant:

(4/10)*(6/9) = 4/15, as in case 2 above.

For K to win in this case, (4/15)*(3/8)*(2/7) = 1/35.

Cases 1, 2, and 3 above are mutually exclusive and independent.

So we can add these probabilities: (1/14) + (1/35) + (1/35) = (1/14) + (2/35) = (5/70) + (4/70) = 9/70.

Do you agree?