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 be my fault.)
(ii) The 3 cards are of different suits.
The first card can be any card: .
. . The second card must be of another suit: .
. . . . The third card must be of yet another suit: .
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): .