Probability Question #2?

P(2K,3A)

= (4C2*48C3 +4C3*48C2 -4C2*4C3) / 52C5

= (6*17296 +4*1128 -6*4) / 2598960

= 108264 / 2598960

= 0.041657

i believe my argument should not be nicely gained. I say the risk is 50% enable a ??, enable b ?? and randomly decide on the values for a and b. As already stated, for a ? 0, P( a < b²) = a million, it is trivial. in elementary terms quite much less trivial is the thought P(a < 0 ) = a million/2 and hence P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what happens whilst a > 0 For a > 0, on the same time because it is trouble-free to coach there's a non 0 danger for a finite b, the shrink, the risk is 0. a < b² is resembling asserting 0 < a < b², bear in mind we are in elementary terms finding at a > 0. If this a finite era on a limiteless line. The danger that a is an ingredient of this era is 0. P( a < b² | a > 0) = 0 As such we've an entire danger P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 bear in mind, this is because of infinite instruments. no count number what form of era you draw on paper or on a working laptop or laptop you will locate a finite danger that seems to attitude a million. yet it is via the finite random style turbines on the laptop and if we had this question asked with finite values there may be a a answer greater effective than 50%. i do no longer advise to be condescending, yet please clarify why using the Gaussian to approximate a uniform distribution is a stunning concept? are not infinite numbers exciting. Cantor whilst mad working with them! :)

kings would be 2/52=1/26 and aces would be 3/52

its 52/5 then 5/2 for kings and 5/3 for aces but i dnot hav a calculator sorry n

4/1

The odds of drawing any one specific card is 1/13.

kings = 1/13 * 1/13 = 1/69

aces = 1/13 * 1/13 * 1/13 = 1/2197

juts multiply 5/52 by 2/5 and get kings...

i think..