For the first bullet point,
My first approach to this problem is as
A bridge hand consists of 13 cards, so that maximum number of pairs that a hand can consist of is 6.
There are 13 ranks in a standard deck of 52 cards.
Examples of different ways to pair (S, H) in a bridge may be the following:
1S 1H, 2S 2H, 3S 3H, 4S 4H, 5S 5H, 6S 6H, 7S or
1S 2H, 2S 3H, 3S 4H, 4S 5H, 5S 6H, 6S 7H, 7S , etc.
Therefore, I believe all possible values of the pair (S, H) would be 13 x 6 = 78.
However, I believe we must be more specific than that and so my second approach is as follows:
There are 4 suites in total and we only want spades and hearts, so there are (4 choose 2) ways to acquire that.
If we only take all the cards of 2 suites, there are 26 cards of those in total and 13 cards to choose for a bridge hand. Thus, there are (26 choose 13) ways to choose the cards in the hand.
Therefore, all possible values of the pair (S, H) would be (4 choose 2) x (26 choose 13) ?
For the second bullet point,
P (S = s, H = h) = P(S=s) and P(H=h).
I am unsure as to how we can generalize the way to find the probabilities of any s and h.
Recently Asked Questions
- how to solve these problems?
- Discrete Probability, Binomial & Poisson 1.Use the following table to calculate the expected value. The following table describes the possible outcomes and
- Analyze the obesity rate of Japan by comparing its obesity rate to the other countries listed in the Global Health Summary data set.Complete the steps below to