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 January 3rd, 2017, 04:09 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 04:24 AM.
 January 4th, 2017, 02:28 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, 07:01 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 121 Thanks: 19 $\displaystyle P(A,B)=\binom{A}{B}$ $\displaystyle P(a)=\frac{P(s)P(l)}{P(A,B)}=\frac{\binom{6}{3} \binom{6}{2}}{\binom{20}{5}}$ Thanks from palongze
January 4th, 2017, 09:06 AM   #4
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 Originally Posted by idontknow $\displaystyle P(A,B)=\binom{A}{B}$ $\displaystyle P(a)=\frac{P(s)P(l)}{P(A,B)}=\frac{\binom{6}{3} \binom{6}{2}}{\binom{20}{5}}$
That the answer for questions (a) i think. Still not clarify about the problems, anyhow thank you for answer.

 January 5th, 2017, 05:48 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, 06:28 PM #6 Senior Member     Joined: Sep 2015 From: CA Posts: 936 Thanks: 506 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 $n_{0l} = {_{14}C}_5 = 2002$ $p_{0l} = \dfrac{2002}{15504} = \dfrac{1001}{7752}$ Thanks from palongze
January 5th, 2017, 07:02 PM   #7
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Quote:
 Originally Posted by romsek sweets arranged as $(s,l,o)=(6,6,$ 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 $n_{0l} = {_{14}C}_5 = 2002$ $p_{0l} = \dfrac{2002}{15504} = \dfrac{1001}{7752}$
Are the answer will be different if the event of picking 5 sweets is done in one move and not one by one. (5 movements)? A bit confused since if we take one sweet at a time, we will pick 5 times versus we take all 5 sweet all together.

January 5th, 2017, 07:32 PM   #8
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Quote:
 Originally Posted by palongze Are the answer will be different if the event of picking 5 sweets is done in one move and not one by one. (5 movements)? A bit confused since if we take one sweet at a time, we will pick 5 times versus we take all 5 sweet all together.
it makes no difference

could you even detect if you picked up the candies one at a time separated by a nanosecond?

January 5th, 2017, 07:39 PM   #9
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Quote:
 Originally Posted by romsek it makes no difference could you even detect if you picked up the candies one at a time separated by a nanosecond?
Noted. Thank you for the explanations, at last my problems has been settled.

### simultaneously and one by one probability

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