Alexane has 12 oreo cookies of which 9 are the regular type and 3 are golden type.
She wants to share her oreos with her 3 friends Jose, Amy, and Camille.
She puts the 12 oreos in a brown bag and she and her 3 friends each randomly choose 3 oreos.
What is the probability that exactly one of them ends up with all 3 golden oreos?
If the probability can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .
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Can you point out the error:
Every golden cookie has 4 options so altogether the golden Oreo arrangement is 4×4×4 = 64...
Out of this 4 are favourable cases..
Hence answer is 4\64 = 1\16
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your answer seems like some ancient probability calculation. eg a lottery ticket would only win or lose the prize, but you cannot say that you have 50% chance to win the prize. You have to add a coefficient to each case. since there are way more chance to distribute the goldens into diffenert guys.
Nicely explained sir!! +1
12C3 is 220 A little typo
Is it wrong if I interpret the following two questions as different problems?
A) we shall distribute 12 cookies (9black+3gold) among four friends, and what is the probability that one among them gets 3 gold ones? I consider your solution to this question.
B) 12 cookies (9black identical +3gold identical) are distributed among four friends. Now what is the probability that one among them has 3 gold cookies?
3 gold cookies may have been distributed among 4 friends in following ways:
Three friends with only one gold each= 4 ways
One friend with one gold, another with 2 golds = 12 ways
One among the four has 3 golds= 4 ways.
Probability= 4/20 = 1/5
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I think it is because each of the WAYS (as you mentioned) do not have the same chance of happening.
For an INDIVIDUAL to have (3 gold , 0 black)=(3 C 3)(9 C 0)=1 , (2 gold, 1 black)= (3 C 2)(9 C 1)=27 ,
(1 gold, 2 black)=(3 C 1)(9 C 2)=108 and (0 gold ,3 black)=(3 C 0)(9 C 3)=84
We have 1+27+108+84=220 combinations and a person getting 1 golden oreo versus getting 3 golden oreos is 108 to 1.
Your solution seems to imply that one among the four has 3 golds= three friends with one gold each... ( 4 ways each)
I think that one having 3 golden oreos is RARER then 3 having 1 each...
As an alternative to Paul's solution (my first preference) here is another approach.
Whoever it is who first chooses a golden Oreo, what matters is that the second and third Oreos she chooses are both golden. The probability that her second Oreo is golden is $\tfrac{2}{11}$ , and the probability that her third Oreo is also golden is then $\tfrac{1}{10}$ . Thus the probability that somebody gets all three golden Oreos is $\frac{2}{11} \times \tfrac{1}{10} = \tfrac{1}{55}$ .
Thanks for this really intuitive solution!
Why is the probability of choosing the first golden oreo not also considered?
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It is not considered, because somebody has to choose the first one and here it does not matter who does, only what matters is she has to end up picking all 3.
This approach makes several unjustified assumptions. First, the probabilities assumed require that the first person picks a golden oreo, since it presumes that there are 11 remaining. It also presumes that the same person gets to make 3 picks in a row. The fact that the answer is correct is doesn't forgive these errors.
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I am afraid you are wrong; it does not matter if I make these assumptions. You seem to be assuming that what Oreo cookies the others have chosen matters. It would, but only if we knew what those choices were. In other words, if each girl ate one cookie, and told everyone what they had, then the problem would indeed be totally changed.
Since this is not the case, where the other cookies are - still in the pile up for grabs or as yet unseen in one of the girls's hands - does not matter. If you calculated the probabilities of the various first cookie choices for each girl, and then calculated the probability of second cookie choices conditional upon all those various first cookie choices, and then calculated the final cookie choice probabilities conditional upon the first and second choice options, this massive calculation would simply collapse to the one above.
Here is another way of looking at this. Let us consider Amy. She picks a cookie. For as long as she does not know what the other girls have chosen, her probability of picking a golden Oreo is $\tfrac{3}{12} = \tfrac14$ . Her probability of picking a second golden Oreo is then $\frac{2}{11}$ , and her probability of her picking a third golden one is then $\tfrac{1}{10}$ , giving her a probability of $\tfrac{1}{220}$ of picking three golden Oreos. The other three girls are thinking exactly the same thing, and their arguments are equally valid. Thus the probability that one girl gets three golden Oreos is $4 \times \tfrac{1}{220} = \tfrac{1}{55}$ . My argument, in addition to the above, noted that since the four girls all had the same probability, we did not have to worry which one of them it was, and therefore did not waste time determining the probability of the first cookie.
Consider a simple example. Deal two cards from the top of a pack. What is the probability that the second card is an Ace? If $A_1$ is the event that the first card is an Ace, and $A_2$ is the event that the second card is an Ace, then $P(A_1 \cap A_2) = \tfrac{4}{52}\times\tfrac{3}{51} \qquad P(A_1^c \cap A_2) \; = \; \tfrac{48}{52}\times\tfrac{4}{51}$ and the sum of these probabilities is $P(A_2) \; = \; \tfrac{4 \times 51}{52 \times 51} = \tfrac{1}{13} \; = \; P(A_1)$ The probability that the second card is an Ace is $\tfrac{1}{13}$ , and it does not matter that the first card might also have been an Ace.
The order of picking does not matter. As a similar example, suppose that $13$ people are each dealt $4$ cards from an ordinary, and randomly shuffled, pack of cards. Is any one person's probability of getting $4$ Aces different from anyone else's? No. Does the probability change if the players are each dealt one card, then each dealt a second card, then a third, then a fourth, or if instead each player is simply given four cards off the top of the deck in turn? No.
Of course, if we were playing poker, and the cards were being dealt face up, then the probabilities for each player would be changing all the time. As I have already said, that consideration is not relevant here.
Instead of pulling from a bag, let us say that we are placing them on a table and shuffle them uniformly in random.
In how many ways can I permute all the cookies? $12!$
In how many ways can I pick the person who has the three gold cookies? $4$
So, the probability required is the ratio of the favorable permutations to the total number of permutations, which is $\frac{4 \times 9! \times 3!}{12!}$
Simple standard approach.
Let $A$ be the probability of finding $3$ golden oreos for the first person, $B$ the probability for the second person and so on. We want to find $P(A U B U C U D)$ which is $P(A) + P(B) + P(C) + P(D)$ and, since the probability of each person to win is the same, the probability that one person can find $3$ golden oreos is $4*P(A)$ .
Each person takes $3$ oreos, and there are ${12 \choose 3}$ possible ways to grab three oreos. The only way of finding three oreos that we are interested in is to find three golden oreos, so $P(A) = \frac{1}{{12 \choose 3}}$ which is $\frac{1}{20 * 11}$
To find the possibility that one of the four people finds three golden oreos we have to multiply $P(A)$ for $4$ and we get $\frac{4}{20*11}$ which is $\frac{1}{55}$ .
Then the answer is $56$ .
Considering each cookie as a unique one, we have 12! ways of distributing them.
Among these 12! distributions, 4 × 3! × 9! = 4 × 6 × 9! have the 3 golden cookies together with a single friend.
Then, the probability is:
(4 × 6 × 9!)/12! =
24/(12 × 10 × 11) =
1/55
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We have 12 C 3 = 220 different combinations of 3 oreo cookies.
Receiving all 3 golden oreos is ONE of these combinations that are all equally probable. The probability that a person receives 3 golden oreos is then 1/220.
The probability that any one of the 4 receives 3 golden oreos is 1/220 + 1/220 + 1/220 + 1/220= 1/55 because the events are disjoint .
1+55=56