Respuesta :
Answer:
a) 0.000394
b) 0.0256
c) 0.00002
Step-by-step explanation:
There are C(52,5) ways of selecting 5 cards from a deck of 52, where C(52,5) is combinations of 52 taken 5 at a time.
[tex]C(52,5)=\binom{52}{5}= \frac{52!}{5!(52-5)!}=\frac{52!}{5!47!}=\frac{52.51.50.49.48}{5.4.3.2}=2,598,960[/tex]
a)
There are only four cards ten. Once you have selected a ten, there are only 4 cards 9, 4 cards 8 and so on.
By the principle of counting (rule of product), there are 4*4*4*4*4= 1024 ways of having a straight with a higher card 10. So, the probability of this event is
1,024/2,598,960 = 0.000394
b)
There are 52 ways of choosing a random card. Once that card is chosen, it could be the 1st, 2nd ,3rd ,4th or 5th in the straight.
For each of this possibilities,since poker cards have 4 different kinds, by the rule of product, there are
[tex]52*4^4[/tex] possibles straights, so there are [tex]5*52*4^4[/tex] ways of having a straight. The probability of a 5-card straight is
[tex]\frac{5.52.4^4}{2,598,960}=\frac{66,560}{2,598,960}=0.0256=2.56\%[/tex]
c)
There are only 52 ways of being dealt a straight flush, so the probability of this rare event is
52/2,598,960 = 0.00002
You can use the rule of product to get the total count and thus, the probability for the given cases.
The answers are
a) The probability that it will be straight with high card 10 is 0.00039
b) The probability that it will be a straight is 0.0043
c) The probability that it will be a straight flush is 0.0000169
What is the rule of product in combinatorics?
If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in [tex]p \times q[/tex] ways.
Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.
Thus, doing A then B is considered same as doing B then A
Using the above fact to get the needed probabilities
There is a fact that probability in non bias elementary events is ratio of counts of favorable events to the count of total events.
Evaluating the probability as per the cases,
a) Since the high card is 10 and a straight contains adjacent card, thus, we need other cards as 9,8,7, and 6
Total ways of getting 5 cards from a deck(of 52 cards) = [tex]^{52}C_5 = \dfrac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2598960[/tex]
Taking out 6,7,8,9,10 out of the deck:
there are four copies of each of 6,7,8,9, and 10. Thus, four four ways for each digits' card withdrawing.
Thus, using rule of product from combinatorics, total ways of getting 10,9,8,7,6 cards is [tex]4 \times 4 \times 4 \times 4\times 4 = 4^5 = 1024[/tex]
Thus, we have
The probability that it will be straight with high card 10 is = 1024/2598960 = 0.00039
b) Since a straight with size five can be any 5 adjacent pair from any suit, we can tag each pair with its one element as other 4 will be adjacent to it.
Let we tag each of them with their low card.
Let the low be ace
Then there are, as discussed in case a), has 1024 ways.
Same for low = 1, low = 2, ... , low = 10 (the max straight will be 10, jack, queen, king, ace as ace can be high or low).
Thus, there are total 11 straights (differentiated with digits or ace, and not suit) and each can be taken in 1024 ways, thus, by product rule, their total count is [tex]1024 \times 11 = 11264[/tex] ways
Thus, probability that it will be a straight is = 11264/2598960 = 0.0043
c) Since now there are no 4 copies as in case b, thus we have only 11 straights possible from each suit which makes 11 times 4 = 44 straights.
Its probability is 44/2598960 = 0.0000169
Thus,
The answers are
a) The probability that it will be straight with high card 10 is 0.00039
b) The probability that it will be a straight is 0.0043
c) The probability that it will be a straight flush is 0.0000169
Learn more about probability here:
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