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January 3rd, 2017, 04:09 AM   #1
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Smile 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.
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January 4th, 2017, 02:28 AM   #2
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Can anyone help me? Already ask at many forum but it seems doesn't attractive anyone to answer it.
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January 4th, 2017, 07:01 AM   #3
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$\displaystyle P(A,B)=\binom{A}{B}$
$\displaystyle P(a)=\frac{P(s)P(l)}{P(A,B)}=\frac{\binom{6}{3} \binom{6}{2}}{\binom{20}{5}}$
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January 4th, 2017, 09:06 AM   #4
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Originally Posted by idontknow View Post
$\displaystyle P(A,B)=\binom{A}{B}$
$\displaystyle P(a)=\frac{P(s)P(l)}{P(A,B)}=\frac{\binom{6}{3} \binom{6}{2}}{\binom{20}{5}}$
That the answer for questions (a) i think. Still not clarify about the problems, anyhow thank you for answer.
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January 5th, 2017, 05:48 PM   #5
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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.
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January 5th, 2017, 06:28 PM   #6
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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

$n_{0l} = {_{14}C}_5 = 2002$

$p_{0l} = \dfrac{2002}{15504} = \dfrac{1001}{7752}$
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January 5th, 2017, 07:02 PM   #7
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Quote:
Originally Posted by romsek View Post
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.
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January 5th, 2017, 07:32 PM   #8
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Quote:
Originally Posted by palongze View Post
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?
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January 5th, 2017, 07:39 PM   #9
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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.
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