Probability of two random selections
$begingroup$
A person has 2 assignments:
- Randomly select 2 best performing stocks of 3 stocks.
- 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:
Prob = $frac{1}{3}$
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
$endgroup$
add a comment |
$begingroup$
A person has 2 assignments:
- Randomly select 2 best performing stocks of 3 stocks.
- 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:
Prob = $frac{1}{3}$
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
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
A person has 2 assignments:
- Randomly select 2 best performing stocks of 3 stocks.
- 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:
Prob = $frac{1}{3}$
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
$endgroup$
A person has 2 assignments:
- Randomly select 2 best performing stocks of 3 stocks.
- 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:
Prob = $frac{1}{3}$
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
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:)
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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:)
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:)
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:)
$endgroup$
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:)
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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