The bucket containing orange tennis balls and yellow tennis balls from which 5 balls are selected at random, but assume that the bucket contains 9 orange balls and 7 yellow balls. (1) what is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow? (1) what is the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow?

(1)  

Probability that all are yellow =

Probability that all are orange =

Sum of the two probabilities = 0048+ 0.0288=0.0336

Thus, the probability that at least one is orange and at least one is yellow = 1-0.0336= 0.9664= 96.64 %

(2) In addition to the above, we eliminate the "1 and 4" cases.  

In these cases, any of the 5 can be taken out, so we multiply by 5.  

1 yellow + 4 orange =

1 orange + 4 yellow =

Sum of two probabilities = 0.2019+0.0721=0.2740  

Now adding 0.0336 to this probability =0.0336+0.2740= 0.3076  

Subtracting that from 1.0000 = 1-0.3076=0.6924= 69.24 %

Thus, the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow= 69.24%

c) the median for town a 30" is greater than the median for town b 25"

step-by-step explanation:

town a, median = q2 = 30

town b, median - q2 = 25

so the median for town a 30" is greater than the median for town b 25"