Hello, thesocialnetwork!

Three cards are drawn from an ordinary pack of 52 cards, at random without replacement.

Cards drawn have the following values:

. . Aces score 1 point,

. . Tens, Jacks, Queens and Kings score 10 points,

. . Other number cards score as many points as their numbers.

Find the probabilities that:

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

iii) total score of the 3 cards is more than 28, given that all 3 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!

I don't agree with any of the answers.

(But that could befault.)my

(ii) The 3 cards are of different suits.

The first card can becard: .any

. . The second card must be of another suit: .

. . . . The third card must be of yet another suit: .

(iii)

The numerator has two cases:

. .

. . . . There are orders.

. . . . The first card can be any Nine: .

. . . . . The second is a Ten of another suit: .

. . . . . . The third is a Ten of yet another suit: .

. . Hence: .

. . . (Of course, the 10's are of different suits!)

. . . . There are choices.

. . . . The first card can be any Ten: .

. . . . The second card can be any other Ten: .

. . . . The third card can be any other Ten: .

. . Hence: .

The numerator is: .

The denominator was determined in part (ii): .

Therefore: .