MATH SOLVE

5 months ago

Q:
# 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?

Accepted Solution

A:

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%.

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%.