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 April 20th, 2010, 06:09 PM #1 Newbie   Joined: Apr 2010 Posts: 21 Thanks: 0 Confused A bag contains 30 marbles with 18 blue and 12 green ones. If you reach into the bag and draw out marbles one at the time until you have three in your hand, what is the probability that they are all green? Question 18 answers A. 68/375 B. 11/203 C. 204/1015 D. 27/125 A bag contains five red, three white, and two blue balls. If three balls are selected at random, with replacement, find the probability that they are all red. Question 19 answers A. 1/24 B. 1/8 C. 1/12 D. 1/60 A bag contains five red, three white, and two blue balls. If three balls are selected at random, with replacement, find the probability that the first two are red and the third one is blue. Question 20 answers A. 1/40 B. 1/20 C. 5/72 D. 60
April 20th, 2010, 06:59 PM   #2
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Re: Confused

Hello, katfisher1208!

Quote:
 18. A bag contains 30 marbles: 18 blue and 12 green. If you reach into the bag and draw 3 marbles without replacement, what is the probability that they are all green? [color=beige]. . [/color]$(A)\;\frac{68}{375} \;\;\;\;(B)\;\frac{11}{203} \;\;\;\;(C)\;\frac{204}{1015}\;\;\;\;(D)\;\frac{27 }{125}$

$P(\text{3 green}) \;=\;\frac{12}{30}\,\cdot\,\frac{11}{29}\,\cdot\,\ frac{10}{28} \;=\;\frac{11}{203}\;\;(B)$

Quote:
 19. A bag contains: 5 red, 3 white, and 2 blue balls. If 3 balls are selected at random, with replacement, find the probability that they are all red. [color=beige]. . [/color]$(A)\;\frac{1}{24} \;\;\;\;(B)\;\frac{1}{8} \;\;\;\;(C)\;\frac{1}{12} \;\;\;\;(D)\;\frac{1}{60}$

$P(\text{red}) \:=\:\frac{5}{10}\:=\:\frac{1}{2}$

$\text{Therefore: }\;P(\text{red, red, red}) \;=\;\frac{1}{2}\,\cdot\,\frac{1}{2}\,\cdot\,\frac {1}{2} \;=\;\frac{1}{8}\;\;(B)$

Quote:
 20. A bag contains: 5 red, 3 white, and 2 blue balls. If 3 balls are selected at random, with replacement, find the probability that the first two are red and the third one is blue. [color=beige]. . [/color]$(A)\;\frac{1}{40} \;\;\;(B)\;\frac{1}{20} \;\;\;(C)\;\frac{5}{72} \;\;\;(D)\;60$

$P(\text{red}) \:=\:\frac{5}{10} \:=\:\frac{1}{2} \;\;\;\;\;{(\text{blue}) \:=\:\frac{2}{10} \:=\:\frac{1}{5}$

$\text{Therefore: }\;P(\text{red, red, blue}) \;=\;\frac{1}{2}\,\cdot\,\frac{1}{2}\,\cdot\,\frac {1}{5} \;=\;\frac{1}{20}\;\;(B)$

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