Answer:
A) i) the probability it is brown = 60%. (ii)The probability it is yellow or blue = 25% (iii) The probability it is not green = 90% (iv)The probability it is striped =0%
B) i)The probability they are all brown = 21.6%. (ii) Probability the third one is the first one that is red = 4.51% (iii) Probability none are yellow = 51.2% (iv) Probability at least one is green = 27.1%
Explanation:
A) The probability that it is brown is the percentage of brown we have. However, Brown is not listed, so we subtract what we are given from 100%. Thus;
100 - (20 + 5 + 5 + 10) = 100 - (40) = 60%.
The probability that one drawn is yellow or blue would be the two percentages added together: 20% + 5% = 25%.
The probability that it is not green would be the percentage of green subtracted from 100: 100% - 10% = 90%.
Since there are no striped candies listed, the probability is 0%.
B) Due to the fact that we have an infinite supply of candy, we will treat these as independent events.
Probability of all 3 being brown is found by taking the probability that one is brown and multiplying it 3 times. Thus;
The percentage of brown candy is 60% from earlier. Thus probability of all 3 being brown is;
0.6 x 0.6 x 0.6 = 0.216 = 21.6%
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% - 5% = 95% = 0.95
Now, for the first two and the probability that it is red = 5% = 0.05
Thus for the last being first one to be red = 0.95 x 0.95 x 0.05 = 0.0451 = 4.51%.
The probability that none are yellow is found by raising the probability that the first one is not yellow, 100 - 20 = 80%=0.80, to the third power:
0.80³ = 0.512 = 51.2%.
The probability that at least one is green is; 1 - (probability of no green).
We first find the probability that all three are NOT green:
0.90³ = 0.729
1 - 0.729 = 0.271 = 27.1%.