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Statistics probability help needed (permutation?) 
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#1
Sep3010, 06:26 PM

P: 8

1. The problem statement, all variables and given/known data
An analyst is presented with lists of four stocks and six bonds. He is asked to predict, in order, the two stocks that will yield the highest return over the next year and the two bonds that will have the highest return over the next year. Suppose that these predictions are made randomly and independently of each other. What is the probability that the analyst will be successful in at least one of the two tasks? 3. The attempt at a solution So first off, I attempted to do this problem by interpreting what the "two tasks" meant, which I thought was (1) predict stock successfully and (2) predict bonds successfully. so I think the probability that the analyst predicts the stocks correctly (P(A)) is (2/4)*(1/3)= 1/6, since there is a chance of predicting 2 stocks right out of 4, and then 1 stock right out of the 3 remaining ones. Like predicting stocks, the I think the probability of predicting bonds correctly (P(B)) is: (2/6)*(1/5)= 1/15 Since the events are not mutually exclusive, in order to find the probability that the analyst will be successful in at least one of the two tasks, you can't just add P(A) + P (B)... you would subtract P (A "intersect" B), or the probability ("analyst predicts both successfully"). The problem is, I am not sure how to find P(analyst predicts both successfully), since the stocks and bonds are different events and cannot be calculated through permutation? Otherwise, I suppose I would subtract P(A) + P(B) by (1/6)*(1/15)??? Or would it be easier to find the probability that the analyst predicts NONE of the stocks/bonds correctly? And how would that be calculated? Please let me know if I am going in the right direction, and how to continue solving this question, thank you. 


#2
Oct110, 09:55 AM

P: 319

I don't think your way of calculating the prob's of the tasks ist correct
First the chance of picking two stocks right (neglecting the order) is [tex]P(2,4) = \frac{1}{\begin{pmatrix}4\\2\end{pmatrix}}=\frac{1}{\frac{4!}{2!2!}}=\f rac{1}{6}[/tex] As you stated the question he also has to be right on the order. So you need to find an expression for correctly predicting the first and the second stock/bond. For the final question you are right, that it is easier to calculate the probability for the analyst to predicting none correctly. 


#3
Oct110, 06:58 PM

P: 8

Thanks betel.
Following up with what you said, I tried again, only this time calculating the probability that the analyst predicts none correctly. So, I used a permutation: (I don't know how to write down a permutation in this forum, but imagine for now that the subscript and superscript are directly vertically lined) Stocks: P(A^{c}) = P^{2}_{2} / P^{4}_{2}= 1/6 Bonds P(B^{c}) = P^{4}_{2} / P^{6}_{2} = 6/15 So the complements (probability of predicting incorrectly) are stocks = 1/6, bonds (6/15). What I am having trouble conceptualizing is how to figure out the probability that the analyst predicts neither stocks nor bonds correctly. Is it the union of A^{c} and B^{c}? Would you calculate (1/6) * (6/15) to get the complement? I know P(AUB) = P(A) + P(B)  P(AnB), but in this instance how would you even go about finding P(AnB)? I appreciate anyone's help. 


#4
Oct110, 08:44 PM

P: 8

Statistics probability help needed (permutation?)



#5
Oct210, 05:46 AM

P: 319

The numerator is one because there are [tex]\begin{pmatrix} 4\\2\end{pmatrix} possible different pairs of highest return stocks. As the analyst can only predict one pair and only one pair will be the right one, the numerator has to be 1.
I dont understand how you calculate the probability for the bonds. Why do you take 2 out of 4 first? To calculate the probability of not correctly predicting anything, first write down the prob for not correctly predicting each of the results. The stocks and bonds should develop independently. So there should be no overlap. Remember: My prob was given neglecting the order. For your question you have to include that, so you have to think how it will change the prob. 


#6
Oct210, 10:14 AM

P: 8

Okay, I see what you mean, that definitely makes more sense than whatever I was trying to do.
So similarly, the probability of the analyst predicting the right bond would be 1/ P^{6}_{2} , since he can only predict one pair and only one pair will be the right one, thus, P(predict bonds correctly) = 1/30 The method that you used neglected order, but doesn't the use of permutation take care of the order issue? If I had used the combination method, then that would be discounting order, doesn't it? And to get the probability of not correctly predicting anything, well, the equation is: 1P(A) = P(A^{c}) , if A is the event that the analyst predicts the stocks correctly. P(A^{c}) = 5/6 Probability of not predicting bonds correct: P(B^{c)}) = 29/30 Since I'm looking for the probability of predicting both incorrectly, and events independent, then P(A^{c}nB^{c})= P(A^{c})*P(B^{c}) = 29/36 And since I'm finding the probability that the analyst will be successful in at least one of the two tasks, then I'm just finding the complement of predicting both incorrectly, 1 (29/36). I'm wondering if that makes any sense? 


#7
Oct210, 10:20 AM

P: 319

Your calculation is correct.
But the permutation does not take care of the order. More precisely the permutation eliminates the order. For example if there are only three choices: The probability to pick the first two (w/o order) is 1/3. To pick the first two in the correct order is 1/6. Try to think what combinations are possible then you will find the correct formula for the correct order. 


#8
Oct210, 10:28 AM

P: 8

Thank you betel, you've really been a big help! I think I'm starting to get the hang of these kinds of probability questions.



#9
Oct210, 10:33 AM

P: 319

Welcome. So what answer do you finally get?



#10
Oct210, 10:38 AM

P: 8

P(analyst predicts at least one of the tasks correctly) = 1 (29/36) = 7/36



#11
Oct210, 10:55 AM

P: 319

No, this is the result without the order. If that is what you want it's fine, but in your question you stated



#12
Oct210, 11:10 AM

P: 8

Okay, going back to this example:
P^{3}_{1} Because you have 3 choices, and you're picking the 1st one And to pick the second, you are left with only 2 choices from the 3, so picking one out of that, you have the combination of P^{2}_{1} Because only one choice is correct, you have 1/ 3*2= 1/6 Is that the reasoning behind finding the combinations of picking things in order? 


#13
Oct210, 11:12 AM

P: 319

Correct.



#14
Oct210, 11:21 AM

P: 8

so then you apply the same principles to the stock/bond question,
For stocks: total combinations> P^{4}_{1} * P^{3}_{1} (1/12) is the Probability that the analyst predicts the stocks correctly, in order. For bonds Total combinations > P^{6}_{1} * P^{5}_{1} = 30 (1/30) probability that analyst predicts the bonds correctly, in order Then, after finding the complements of both probabilities, the probability that the analyst predicts both incorrectly is (11/12) * (29/30) = 319/360 is the probability that the analyst predicts both incorrectly, in order Take the complement of that to find the probability that the analyst predicts at least one task correctly 1 (319/360) = 41/360 


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