Apet store has 11 puppies, including 3 poodles, 5 terriers, and 3 retrievers. if rebecca and aaron, in that order, each select one puppy at random without replacement, find the probability that both select a poodle. how would this situation change if the selection was made with replacement?

Hello mate!  

in the store, there are 11 puppies, and 3 of them are poodles.

If Rebbeca and Aaron chose at random, then the probabilities are; for Rebbeca, there is a 3/11 probability to get a poodle ( because there are 3 poodles of eleven dogs) and the probability for Aaron is 2/10 because now there are one less poodle (and in consequence one less puppy)

now, the joint probability of these two events happening is the product of each probability, this is:

p = (3/11)*(2/10) = 6/110, that we can simplify at 3/55 if we divide both denominator and numerator by 2.

If the dog is replaced after Rebbeca picked hers, then Aaron also has a 3/11 probability of picking a poodle, and in that case, the joint probability is:

q = (3/11)*/3/11) = 9/121

now, p = 3/55 = 0.055 and q = 9/121 = 0.074

this means that the probability of both of them selecting a poodle is bigger when the dog is replaced (which makes a lot of sense)

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.