please explain how to resolve all question in part B) For a certain candy, 15% o
ID: 3178428 • Letter: P
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please explain how to resolve all question in part B)
For a certain candy, 15% of the pieces are yellow, 15% are red, 10% are blue, 10% 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? a) The probability that it is brown is 0.50. (Round to three decimal places as needed.) The probability that it is yellow or blue is 0.25. (Round to three decimal places as needed.) The probability that it is not green is 0.90. (Round to three decimal places as needed.) The probability that it is striped is 0. (Round to three decimal places as needed.) b) The probability of picking three brown candies is 0.125 (Round to three decimal places as needed.) The probability of the third one being the first red one is 0.108. (Round to three decimal places as needed.) The probability that none are yellow is 0.614. (Round to three decimal places as needed.) The probability of at least one green candy is 0.271. (Round to three decimal places as needed.)Explanation / Answer
let us assume there are 100 candies. So 15 are Yellow, 15 are red, 10 are blue, 10 are green and the remaining(100-15+15+10+10) = 50are brown.
Please remember P(of an event) = no of favourable outcomes/Total no of outcomes
a) On one pick , that it is a brown = 50/100 = 0.5
The probability it is yellow or blue = (15+10)/100= 25/100 = 0.25
The probability that it is NOT green= 1-(P(of picking green) = 1-(10/100) = 90/100 = 0.9
The probability that it is striped---Here no data is given about striped candies, and hence the probability is 0.
b) Probability of picking 3 brown candies= (0.5*0.5*0.5) = 0.125
Probability of the third being the first red.---here other than red there are 85 other candies. So the first 2 should be from this lot, whose probability = 0.85. therefore required probability is 0.85*0.85*0.15 = 0.108
The probability that none are yellow---here again 15 are yellow, and we should not pick any of them...we need to pick from the other 85 whose probability of being picked is 0.85. therefore the required probability is 0.85*0.85*0.85 = 0.614
Probability of at least 1 green candy = P(1 green, 2 others) + P(2 greens, 1 other) + P(all three are green) =
(0.1*0.9*0.9) + (0.1*0.1*0.9) + (0.1*0.1*0.1) = 0.081 + 0.009 + 0.001 = 0.091
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