Question

For a certain​ candy, 15​% of the pieces are​ yellow, 1010​% are​ red, 2020​% are​ blue, 55​% are​ green, and the rest are brown. ​a) if you pick a piece at​ random, what is the probability that it is​ brown? it is yellow or​ blue? it is not​ green? it is​ striped? ​b) assume you have an infinite supply of these candy pieces from which to draw. if you pick three pieces in a​ row, what is the probability that they are all​ brown? the third one is the first one that is​ red? none are​ yellow? at least one is​ green?

Answer 1

A) the probability it is brown would be 50%; the probability it is yellow or blue would be 35%; the probability it is not green is 95%; the probability it is striped is 0%.
B) the probability of all brown would be 12.5%; the probability that the third one is the first red one drawn is 8.1%; the probability that none are yellow is 61.4%; the probability that at least one is green is 14.3%.

Explanation:
A) The probability that it is brown is the percentage of brown we have.  Brown is not listed, so we subtract what we are given from 100%:
100-(15+10+20+5) = 100-(50) = 50%.  The probability that one drawn is yellow or blue would be the two percentages added together:  15+20 = 35%.  The probability that it is not green would be the percentage of green subtracted from 100:  100-5=95%.  Since there are no striped candies listed, the probability is 0%.
B) Since we have an infinite supply of candy, we will treat these as independent events.  All 3 being brown is found by taking the probability that one is brown and multiplying it 3 times:
0.5*0.5*0.5 = 0.125 = 12.5%.  
To find the probability that the first one that is red is the third one drawn, we take the probability that it is NOT red, 100-10 = 90% = 0.9, for the first two, and the probability that it IS red, 10% = 0.1, for the last:
0.9*0.9*0.1 = 0.081 = 8.1%.

The probability that none are yellow is found by raising the probability that the first one is not yellow, 100-15=85%=0.85, to the third power:

0.85^3 = 0.614 = 61.4%.

The probability that at least one is green is computed by subtracting 1-(probability of no green).  We first find the probability that all three are NOT green:
0.95^3 = 0.857375
1-0.857375 = 0.143 = 14.3%.