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January 3rd, 2017, 03:09 AM  #1 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  Probability of Events, Simultaneously or One by One Events
Hi and have a good day. Need help on solving this questions. Adam is given a bag of 20 sweets of which 6 are strawberry flavoured, 6 are lemon flavoured and 8 are orange flavoured. Adam take out 5 sweets at random and eats them. Find the probability that he eats: (a) 3 strawberry flavoured and 2 lemons flavoured (b) exactly 2 strawberry flavoured sweets (c) no lemon flavoured sweets Are there any differences in answer if the event occurred simultaneously? What I mean is "the event Adam take 5 sweets on a single picking" is different than "Adam take 1 sweets and then pick another sweets until 5 sweets" has been pickup?Another things is are there any easier less tedious method for solving this questions as it was event of picking 5 times sweets and assuming Adam take one by one we will have 3^5 branch of tree diagram which is 243? My thoughts: Assuming the events was pick one by one. (a) S,S,S,L,L Find no of ways the letter can be arranged, we get: 5!/(3! x 2!) = 10 ways, thus the probability: 10 x (6/20)(5/19)(4/1(6/17)(5/16) = 25/1292 My answer for questions (a). Maybe correct maybe wrong. Best regards, Faizal. Last edited by palongze; January 3rd, 2017 at 03:24 AM. 
January 4th, 2017, 01:28 AM  #2 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus 
Can anyone help me? Already ask at many forum but it seems doesn't attractive anyone to answer it. 
January 4th, 2017, 06:01 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 157 Thanks: 21 
$\displaystyle P(A,B)=\binom{A}{B}$ $\displaystyle P(a)=\frac{P(s)P(l)}{P(A,B)}=\frac{\binom{6}{3} \binom{6}{2}}{\binom{20}{5}}$ 
January 4th, 2017, 08:06 AM  #4 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  
January 5th, 2017, 04:48 PM  #5 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus 
Does anyone can help me regarding my questions? Or my questions is too easy and not worth to be answered? Probability is really my weakness and I do hope someone in My Math Forum can help me Sincerely need help Best regards. 
January 5th, 2017, 05:28 PM  #6 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,503 Thanks: 758 
sweets arranged as $(s,l,o)=(6,6,8)$ How many ways can we choose 5 sweets? $n={_{20}C}_5 =15504 $ 1) 3s, 2l. $n_{3s2l} = {_{6}C}_3 \cdot {_{6}C}_2 = 300$ $p_{3s2l} = \dfrac{n_{3s2l}}{n} = \dfrac{300}{15504}=\dfrac{25}{1292}$ 2) 2s $n_{2s} = {_{6}C}_2 \cdot {_{14}C}_3 = 5460$ $p_{2s} = \dfrac{5460}{15504} = \dfrac{455}{1292}$ 3) 0l $n_{0l} = {_{14}C}_5 = 2002$ $p_{0l} = \dfrac{2002}{15504} = \dfrac{1001}{7752}$ 
January 5th, 2017, 06:02 PM  #7  
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  Quote:
 
January 5th, 2017, 06:32 PM  #8  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,503 Thanks: 758  Quote:
could you even detect if you picked up the candies one at a time separated by a nanosecond?  
January 5th, 2017, 06:39 PM  #9 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  

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