Probability of two random selections












0












$begingroup$


A person has 2 assignments:




  1. Randomly select 2 best performing stocks of 3 stocks.

  2. Randomly select the best bond and the second best bond of 6 bonds.


Now what is the probability to make a successful random selection for at least one of the two assignments?



And what is the probability that assignment one is successful and assignment 2 is not successful?



My calculations:




  1. Prob = $frac{1}{3}$


  2. Prob = $frac{1}{15}$



Now using venn diagram logic shouldn't the solution be:
$frac{1}{3 }+ frac{1}{15} - frac{1}{3}×frac{1}{15}$



According to the book the answer is
0.355556



Can someone explain this in venn diagram logic?



I'm still struggling with the second question.






probability probability-theory statistics permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe the question asked for the probability that only one of the assignments is correct? That would get the books answer ($frac{16}{45}$)
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:57
















0












$begingroup$


A person has 2 assignments:




  1. Randomly select 2 best performing stocks of 3 stocks.

  2. Randomly select the best bond and the second best bond of 6 bonds.


Now what is the probability to make a successful random selection for at least one of the two assignments?



And what is the probability that assignment one is successful and assignment 2 is not successful?



My calculations:




  1. Prob = $frac{1}{3}$


  2. Prob = $frac{1}{15}$



Now using venn diagram logic shouldn't the solution be:
$frac{1}{3 }+ frac{1}{15} - frac{1}{3}×frac{1}{15}$



According to the book the answer is
0.355556



Can someone explain this in venn diagram logic?



I'm still struggling with the second question.






probability probability-theory statistics permutations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe the question asked for the probability that only one of the assignments is correct? That would get the books answer ($frac{16}{45}$)
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:57














0












0








0





$begingroup$


A person has 2 assignments:




  1. Randomly select 2 best performing stocks of 3 stocks.

  2. Randomly select the best bond and the second best bond of 6 bonds.


Now what is the probability to make a successful random selection for at least one of the two assignments?



And what is the probability that assignment one is successful and assignment 2 is not successful?



My calculations:




  1. Prob = $frac{1}{3}$


  2. Prob = $frac{1}{15}$



Now using venn diagram logic shouldn't the solution be:
$frac{1}{3 }+ frac{1}{15} - frac{1}{3}×frac{1}{15}$



According to the book the answer is
0.355556



Can someone explain this in venn diagram logic?



I'm still struggling with the second question.






probability probability-theory statistics permutations







share|cite|improve this question











$endgroup$




A person has 2 assignments:




  1. Randomly select 2 best performing stocks of 3 stocks.

  2. Randomly select the best bond and the second best bond of 6 bonds.


Now what is the probability to make a successful random selection for at least one of the two assignments?



And what is the probability that assignment one is successful and assignment 2 is not successful?



My calculations:




  1. Prob = $frac{1}{3}$


  2. Prob = $frac{1}{15}$



Now using venn diagram logic shouldn't the solution be:
$frac{1}{3 }+ frac{1}{15} - frac{1}{3}×frac{1}{15}$



According to the book the answer is
0.355556



Can someone explain this in venn diagram logic?



I'm still struggling with the second question.






probability probability-theory statistics permutations




probability probability-theory statistics permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 0:32









Martund

1,667213




1,667213










asked Dec 31 '18 at 0:01









Lime3645Lime3645

162




162












  • $begingroup$
    Maybe the question asked for the probability that only one of the assignments is correct? That would get the books answer ($frac{16}{45}$)
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:57


















  • $begingroup$
    Maybe the question asked for the probability that only one of the assignments is correct? That would get the books answer ($frac{16}{45}$)
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:57
















$begingroup$
Maybe the question asked for the probability that only one of the assignments is correct? That would get the books answer ($frac{16}{45}$)
$endgroup$
– Zachary Hunter
Dec 31 '18 at 0:57




$begingroup$
Maybe the question asked for the probability that only one of the assignments is correct? That would get the books answer ($frac{16}{45}$)
$endgroup$
– Zachary Hunter
Dec 31 '18 at 0:57










1 Answer
1






active

oldest

votes


















0












$begingroup$

The given solution is correct. The Venn diagram logic is actually Inclusion-Exclusion principle, which says
$$P(Acup B)=P(A)+P(B)-P(Acap B)$$
If events of successful random selection of first and second assignments are $A$ and $B$ respectively, then $$P(A)=frac{1}{binom{3}{2}}=frac{1}{3}$$
$$P(B)=frac{1}{6}×frac{1}{5}=frac{1}{30}$$
You are taking value of $P(B)$ to be $frac{1}{15}$, but there will be an additional factor of $frac{1}{2}$ because the question says," randomly select the best bond and the second best bond of 6 bonds" indicating that there is a possibility of arrangement i.e. the person first chooses the best ball among the 6 balls and then chooses the best ball among the remaining 5 balls. Now, as the assignments are independent, we have,
$$P(Acap B)=P(A)P(B)=frac{1}{3}×frac{1}{30}=frac{1}{90}$$
Now, apply inclusion-exclusion principle and get the result.
$$P(Acup B)=frac{1}{3}+frac{1}{30}-frac{1}{90}=frac{16}{45}$$
For the second question, again use independence.
$$P(Acap B^c)=P(A)P(B^c)=frac{1}{3}×frac{29}{30}=frac{29}{90}$$



Hope it is helpful:)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wouldn't $P(B) = frac{1}{binom{6}{2}}$?
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:56










  • $begingroup$
    No, that is the mistake done by @Lime3645, there are two cases after selecting 2 balls randomly, the person has to declare which one is the best and which is the second best.
    $endgroup$
    – Martund
    Dec 31 '18 at 0:59










  • $begingroup$
    Hm... I am suspicious that it was miscopied. After all, to say "of 6 bonds" the second time is unnecessarily confusing.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 1:03











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

The given solution is correct. The Venn diagram logic is actually Inclusion-Exclusion principle, which says
$$P(Acup B)=P(A)+P(B)-P(Acap B)$$
If events of successful random selection of first and second assignments are $A$ and $B$ respectively, then $$P(A)=frac{1}{binom{3}{2}}=frac{1}{3}$$
$$P(B)=frac{1}{6}×frac{1}{5}=frac{1}{30}$$
You are taking value of $P(B)$ to be $frac{1}{15}$, but there will be an additional factor of $frac{1}{2}$ because the question says," randomly select the best bond and the second best bond of 6 bonds" indicating that there is a possibility of arrangement i.e. the person first chooses the best ball among the 6 balls and then chooses the best ball among the remaining 5 balls. Now, as the assignments are independent, we have,
$$P(Acap B)=P(A)P(B)=frac{1}{3}×frac{1}{30}=frac{1}{90}$$
Now, apply inclusion-exclusion principle and get the result.
$$P(Acup B)=frac{1}{3}+frac{1}{30}-frac{1}{90}=frac{16}{45}$$
For the second question, again use independence.
$$P(Acap B^c)=P(A)P(B^c)=frac{1}{3}×frac{29}{30}=frac{29}{90}$$



Hope it is helpful:)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wouldn't $P(B) = frac{1}{binom{6}{2}}$?
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:56










  • $begingroup$
    No, that is the mistake done by @Lime3645, there are two cases after selecting 2 balls randomly, the person has to declare which one is the best and which is the second best.
    $endgroup$
    – Martund
    Dec 31 '18 at 0:59










  • $begingroup$
    Hm... I am suspicious that it was miscopied. After all, to say "of 6 bonds" the second time is unnecessarily confusing.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 1:03
















0












$begingroup$

The given solution is correct. The Venn diagram logic is actually Inclusion-Exclusion principle, which says
$$P(Acup B)=P(A)+P(B)-P(Acap B)$$
If events of successful random selection of first and second assignments are $A$ and $B$ respectively, then $$P(A)=frac{1}{binom{3}{2}}=frac{1}{3}$$
$$P(B)=frac{1}{6}×frac{1}{5}=frac{1}{30}$$
You are taking value of $P(B)$ to be $frac{1}{15}$, but there will be an additional factor of $frac{1}{2}$ because the question says," randomly select the best bond and the second best bond of 6 bonds" indicating that there is a possibility of arrangement i.e. the person first chooses the best ball among the 6 balls and then chooses the best ball among the remaining 5 balls. Now, as the assignments are independent, we have,
$$P(Acap B)=P(A)P(B)=frac{1}{3}×frac{1}{30}=frac{1}{90}$$
Now, apply inclusion-exclusion principle and get the result.
$$P(Acup B)=frac{1}{3}+frac{1}{30}-frac{1}{90}=frac{16}{45}$$
For the second question, again use independence.
$$P(Acap B^c)=P(A)P(B^c)=frac{1}{3}×frac{29}{30}=frac{29}{90}$$



Hope it is helpful:)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wouldn't $P(B) = frac{1}{binom{6}{2}}$?
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:56










  • $begingroup$
    No, that is the mistake done by @Lime3645, there are two cases after selecting 2 balls randomly, the person has to declare which one is the best and which is the second best.
    $endgroup$
    – Martund
    Dec 31 '18 at 0:59










  • $begingroup$
    Hm... I am suspicious that it was miscopied. After all, to say "of 6 bonds" the second time is unnecessarily confusing.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 1:03














0












0








0





$begingroup$

The given solution is correct. The Venn diagram logic is actually Inclusion-Exclusion principle, which says
$$P(Acup B)=P(A)+P(B)-P(Acap B)$$
If events of successful random selection of first and second assignments are $A$ and $B$ respectively, then $$P(A)=frac{1}{binom{3}{2}}=frac{1}{3}$$
$$P(B)=frac{1}{6}×frac{1}{5}=frac{1}{30}$$
You are taking value of $P(B)$ to be $frac{1}{15}$, but there will be an additional factor of $frac{1}{2}$ because the question says," randomly select the best bond and the second best bond of 6 bonds" indicating that there is a possibility of arrangement i.e. the person first chooses the best ball among the 6 balls and then chooses the best ball among the remaining 5 balls. Now, as the assignments are independent, we have,
$$P(Acap B)=P(A)P(B)=frac{1}{3}×frac{1}{30}=frac{1}{90}$$
Now, apply inclusion-exclusion principle and get the result.
$$P(Acup B)=frac{1}{3}+frac{1}{30}-frac{1}{90}=frac{16}{45}$$
For the second question, again use independence.
$$P(Acap B^c)=P(A)P(B^c)=frac{1}{3}×frac{29}{30}=frac{29}{90}$$



Hope it is helpful:)






share|cite|improve this answer









$endgroup$



The given solution is correct. The Venn diagram logic is actually Inclusion-Exclusion principle, which says
$$P(Acup B)=P(A)+P(B)-P(Acap B)$$
If events of successful random selection of first and second assignments are $A$ and $B$ respectively, then $$P(A)=frac{1}{binom{3}{2}}=frac{1}{3}$$
$$P(B)=frac{1}{6}×frac{1}{5}=frac{1}{30}$$
You are taking value of $P(B)$ to be $frac{1}{15}$, but there will be an additional factor of $frac{1}{2}$ because the question says," randomly select the best bond and the second best bond of 6 bonds" indicating that there is a possibility of arrangement i.e. the person first chooses the best ball among the 6 balls and then chooses the best ball among the remaining 5 balls. Now, as the assignments are independent, we have,
$$P(Acap B)=P(A)P(B)=frac{1}{3}×frac{1}{30}=frac{1}{90}$$
Now, apply inclusion-exclusion principle and get the result.
$$P(Acup B)=frac{1}{3}+frac{1}{30}-frac{1}{90}=frac{16}{45}$$
For the second question, again use independence.
$$P(Acap B^c)=P(A)P(B^c)=frac{1}{3}×frac{29}{30}=frac{29}{90}$$



Hope it is helpful:)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 0:54









MartundMartund

1,667213




1,667213












  • $begingroup$
    Wouldn't $P(B) = frac{1}{binom{6}{2}}$?
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:56










  • $begingroup$
    No, that is the mistake done by @Lime3645, there are two cases after selecting 2 balls randomly, the person has to declare which one is the best and which is the second best.
    $endgroup$
    – Martund
    Dec 31 '18 at 0:59










  • $begingroup$
    Hm... I am suspicious that it was miscopied. After all, to say "of 6 bonds" the second time is unnecessarily confusing.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 1:03


















  • $begingroup$
    Wouldn't $P(B) = frac{1}{binom{6}{2}}$?
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 0:56










  • $begingroup$
    No, that is the mistake done by @Lime3645, there are two cases after selecting 2 balls randomly, the person has to declare which one is the best and which is the second best.
    $endgroup$
    – Martund
    Dec 31 '18 at 0:59










  • $begingroup$
    Hm... I am suspicious that it was miscopied. After all, to say "of 6 bonds" the second time is unnecessarily confusing.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 1:03
















$begingroup$
Wouldn't $P(B) = frac{1}{binom{6}{2}}$?
$endgroup$
– Zachary Hunter
Dec 31 '18 at 0:56




$begingroup$
Wouldn't $P(B) = frac{1}{binom{6}{2}}$?
$endgroup$
– Zachary Hunter
Dec 31 '18 at 0:56












$begingroup$
No, that is the mistake done by @Lime3645, there are two cases after selecting 2 balls randomly, the person has to declare which one is the best and which is the second best.
$endgroup$
– Martund
Dec 31 '18 at 0:59




$begingroup$
No, that is the mistake done by @Lime3645, there are two cases after selecting 2 balls randomly, the person has to declare which one is the best and which is the second best.
$endgroup$
– Martund
Dec 31 '18 at 0:59












$begingroup$
Hm... I am suspicious that it was miscopied. After all, to say "of 6 bonds" the second time is unnecessarily confusing.
$endgroup$
– Zachary Hunter
Dec 31 '18 at 1:03




$begingroup$
Hm... I am suspicious that it was miscopied. After all, to say "of 6 bonds" the second time is unnecessarily confusing.
$endgroup$
– Zachary Hunter
Dec 31 '18 at 1:03


















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