thesocialnetwork wrote:

Three cards are drawn from an ordinary pack of 52 cards, at random and without replacement. Cards drawn have the following values:

Aces score 1 point,

Tens, Jacks, Queens and Kings score 10 points and

Cards from two to nine score as many points as the numbers they carry (i.e. twos score 2 points, threes score 3 points and so on)

Find the probabilities that

ii) all three cards are of different suits (I found this, it's 197/425)

iii) total score of the three cards is more than 28 given that all three cards are of different suits.

what is the answer to iii)? I'm pretty sure my method is right, but I got 160/2197 and the solution given is 112/2197. If you can get 112/2197, please share. Thanks!

My method:

I found P(scores are 10, 10, 9 and of diff suits) + P(scores are 10, 10, 10 and of diff suits) and divided this by P(cards are of different suits). I got 160/2197 but apparently the answer is 112/2197.

Thanks for your help!

You didn't show how you get 160/2197, so I can't pin down where your mistake come from. But part (ii) shouldn't give you a prime numerator (the numerator should be divisible by 13^2, as a check).

Counting the ordered triples:

29 points: T-point Suit #1 (4 ways), T-point Suit #2 (4 ways), 9-point Suit #3 (1 way), 3 permutation of card values (T,T,9), 24 ways to choose the suits (#1,#2,#3), in total: 1152 triples.

30 points: T-point Suit #1 (4 ways), T-point Suit #2 (4 ways), T-point Suit #3 (4 ways), 1 permutation of card values (T,T,T), 24 ways to choose suits, in total: 1536 triples.

So there are 2688 triples good triples, out of 52*39*26, i.e. probability 112/2197 as given.