Question 1:
I am working on probabilities and on some exercises the solutions either use the nCk or just n. I want to find a method to understand when I have to use the nCk and when not to.
Example 1 (nCk is used):
What is the probability that a five-card poker hand contains the two of diamonds, the three of spades, the six of hearts, then ten of clubs, and the king of hearts?
Solution: 1/ 52 C 5Example 2 (nCk is not used):
What is the probability that a five-card poker hand does not contain the queen of hearts?
Solution: 47/52Two similar questions with different Sample Space! Is any tip or standard rule that can be used to determine whether a combination or Permutation must be used, and when just the number of all possibles events (e.g S=52) ?
Question 2:
Some exercises do not determine which method must be used (Combinations or Permutations). As I know, Combinations is used when order doesn't matters and repetition is not allowed, on the other hand Permutations is used when order matters and repetition is not allowed. Both have another kind of method where the repetition is allowed (n+k,C,k and n^k respectively). Any tip or trick when should I use the Combinations or Permutations? and when to use the unlimited repetition for both? I knew that exercise must make it clear, but I find it a bit hard to deal with this choice of the method.
Thank you in advance.
$\endgroup$3 Answers
$\begingroup$You can use even permutations in you first question! Why not? It will be just multiplying $5!$ in both numerator and denominator which just get cancelled out!
Also in the second case : $\dfrac{^{51}C_5}{^{52}C_5}=\frac{47}{52}$
First represents choosing $5$ cards from pack without the heart queen and denominator from all cards.
Both give same result.
And generally the question states what to do. If it is unclear, just do what you feel easier. Because people forget to explain elementary cases. But, still, the question should make it clear.
$\endgroup$ 4 $\begingroup$Combinations are just another name for "set". Order of the elements is irrelevant.
Permutations are arrangements where order is important.
This should cover most cases. Two general rules (from which the above can be derived) is that if you have a sequence of decisions (like selecting each element in a cartesian product of sets) you multiply. If you have a union of disjoint sets, add sizes of the sets to get the total size. Sometimes you are including elements twice, subtract them. It is sometimes easier to divide into a collection of disjoint sets that are easy to handle.
Look if it is easier to use the sum rule backwards: Compute the total, subtract those that don't satisfy the restrictions.
$\endgroup$ 0 $\begingroup$Dr Mathematician P(x)=nCr*(p^r)((q)^(n-r)) is normally used when you have a case of "Success" and "Failure". The formula is P(x)=nCr(p^r)*((q)^(n-r)) In Q1 they really specify 2 of Diamon, 3 of Spades, 6 of Heart, 10 of club and king of heart. So you will be calculating probability of each card, then you will have to sum it up in the end. In this case each card will be of 20%, therefor change to 0.20 and it will be equal to p and q automatically will be equal to 0.80 Q1 they give you five card which is equal to 100%, then they give you Question 1 is a type of Binomial Distribution. This should work. ;-)
Notice nPr is a Permutation Probability Concepts.
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