Choose any two cards, 2, 4, 6, 7, 8, 1 1, 12, 13, and combine the two numbers on these two cards to form a score. Find the probability that a fraction is irreducible (denominator of the numerator is not greater than 1).
Solution 1: (regardless of the combination order of the two numbers obtained)
Because you choose any two cards, the two numbers x and y above form two scores (x, y) or (y, x), and these two scores are either irreducible or irreducible, so they must be one and only one.
Therefore, the combination order of the two numbers can be ignored at this time, that is, (x, y) or (y, x) can be regarded as "a fraction". In this way, the sample space tested here can be clearly expressed as:
ω 1 = {(x，y):y & gt； X = 2，4，6，8， 1 1， 12， 13} = {(2，4)，(2，6)，(2，7)，(2，8)，(2，6438) y=2，4，6，…， 12， 13}={( 13， 12)，( 13
Let Ai={ the score obtained is the converted score} i= 1, 2.
Then a 1≦{(x, y): y >; X = 2, 4, 6, 7, 8, 1 1 2, 13, x and y coprime} = {(2, 7) (2,1/kloc-0. 13)(7,8)(7, 1 1)(7, 12)(7, 13)(8, 1 1)(8, 13)( 1 1, 12)( 1 1, 13)( 12, 13)}? Ω 1
Or a2≦{(x, y): x >; Y = 2, 4, 6, 7, 8, 1 1 2, 13, x and y coprime} = {( 13, 12), (.
Therefore, the number of sample points contained in ω 1 and ω 2 is:
n = nω 1 = nω2 = 7+6+5+4+3+2+ 1 =(7+ 1)72 = 28
A 1 and A2 contain the following sample points: m=mA 1=mA2= 18.
Therefore, from the classical probability calculation formula (1), there are
p(Ai)= Mn = 1828 = 9 14 I = 1，2 .
Scheme 2: (Considering the combination order of the two numbers)
As two cards are randomly selected, the two numbers X and Y on the cards form two scores (X, Y) or (Y, X), which are naturally different. Therefore, as far as the number of different scores is concerned (considering the composition order of the scores obtained by two numbers), the sample space ω tested here can be clearly expressed as ω = ω 1 ∪ ω 2, and the event A for finding the probability can be
Therefore, from the classical probability calculation formula (1), there are
p(A)= nanω= ma 1+ma2nω 1+nω2 = 18+ 1828+28 = 9 14
In this example, we can easily get the number of sample points nω and mA contained in the sample space ω and the event A with our desired probability through the "explicit representation" of the sample space ω, and then get the probability P(A) of the event A with our desired probability by using the classical probability calculation formula (1). However, it should be noted that this method will be effective for some simple test cases, but it will be ineffective for general complex test cases, because its sample space is difficult to be "clearly expressed". For example, in this case, if there are many cards available for withdrawal, there will be many corresponding scores. Writing them down one by one, it is obviously difficult to calculate the total score and the number of points deducted. At this time, it is generally necessary to consider using it.