A box contains 20 red balls, 25 white balls, and 55 blue balls. Suppose that 10 balls are selected at random one at a time with replacement; that is, each selected ball is replaced in the box before the next selection is made. Determine the probability that at least one color will be missing from the 10 selected balls.

0.164 = 16.4% probability that at least one color will be missing from the 10 selected balls.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Determine the probability that at least one color will be missing from the 10 selected balls.

Red missing:

Each time, there are 55 + 25 = 80 non-red balls, out of 100. So, in each of the 10 trials, 80% = 0.8 probability of not picking a red ball. The probability that no red ball is picked is given by:

(0.8)^10 = 0.1074

White missing:

55 + 20 = 75 non-white balls, out of 100, in each trial. The probability that no white ball is picked is given by:

(0.75)^10 = 0.0563

Blue missing:

45 non-blue balls, out of 100. The probability that no blue ball is picked is given by:

(0.45)^10 = 0.0003

Total:

0.1074 + 0.0563 + 0.0003 = 0.164

0.164 = 16.4% probability that at least one color will be missing from the 10 selected balls.

secant-secant formula:

(whole secant) x (external part) =           (whole secant) x (external part)

(8+y-2) *8=                                           (y+4+6) *6

(y+6)*8 = (y+10) *6

distribute

8y+48 = 6y +60

subtract 6y from each side

2y +48 = 60

subtract 48 from each side

2y = 12

divide by 2

y = 6

sv =     (y+4+6)

sv   = 6+4+6

sv = 16