Jelly bean choosing from a jar.
$begingroup$
A certain opaque jar contains $60$ jelly beans — $22$ white, $38$ non-white. What is the probability that the first two jelly beans removed from the jar will be white?
I am wondering about this problem without replacement. So the first white bean gets eaten, then we choose another bean from the jar.
I thought the answer should be $(22/60) cdot (21/59)$, but I was told the answer was a bit more complicated than I originally assumed and involves some combinatorics... but I'd like to know for sure.
probability
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
$endgroup$
add a comment |
$begingroup$
A certain opaque jar contains $60$ jelly beans — $22$ white, $38$ non-white. What is the probability that the first two jelly beans removed from the jar will be white?
I am wondering about this problem without replacement. So the first white bean gets eaten, then we choose another bean from the jar.
I thought the answer should be $(22/60) cdot (21/59)$, but I was told the answer was a bit more complicated than I originally assumed and involves some combinatorics... but I'd like to know for sure.
probability
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
$endgroup$
$begingroup$
I think the issue is how to interpret the question. You are interpreting it as "what's the probability that the first two jelly beans are both white" but I'd say that was not what was intended. Rather, I think the idea was that someone draws them all out one at a time and asks for the probability that at some point two white ones are drawn consecutively.
$endgroup$
– lulu
Dec 3 '18 at 23:37
$begingroup$
You really should ask "the person much smarter than you" about how they interpret this question. I find your solution to be correct according to the wording of your question.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:58
$begingroup$
They only said the answer involves combinatorics. In any case, the question I am interested in is how I am interpreting it (it's academic really), so if my answer seems right in that context, then I'll leave it at that
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:40
add a comment |
$begingroup$
A certain opaque jar contains $60$ jelly beans — $22$ white, $38$ non-white. What is the probability that the first two jelly beans removed from the jar will be white?
I am wondering about this problem without replacement. So the first white bean gets eaten, then we choose another bean from the jar.
I thought the answer should be $(22/60) cdot (21/59)$, but I was told the answer was a bit more complicated than I originally assumed and involves some combinatorics... but I'd like to know for sure.
probability
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
$endgroup$
A certain opaque jar contains $60$ jelly beans — $22$ white, $38$ non-white. What is the probability that the first two jelly beans removed from the jar will be white?
I am wondering about this problem without replacement. So the first white bean gets eaten, then we choose another bean from the jar.
I thought the answer should be $(22/60) cdot (21/59)$, but I was told the answer was a bit more complicated than I originally assumed and involves some combinatorics... but I'd like to know for sure.
probability
probability
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
edited Dec 6 '18 at 19:44
Elijah Rockers
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
asked Dec 3 '18 at 23:00
Elijah RockersElijah Rockers
1014
1014
$begingroup$
I think the issue is how to interpret the question. You are interpreting it as "what's the probability that the first two jelly beans are both white" but I'd say that was not what was intended. Rather, I think the idea was that someone draws them all out one at a time and asks for the probability that at some point two white ones are drawn consecutively.
$endgroup$
– lulu
Dec 3 '18 at 23:37
$begingroup$
You really should ask "the person much smarter than you" about how they interpret this question. I find your solution to be correct according to the wording of your question.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:58
$begingroup$
They only said the answer involves combinatorics. In any case, the question I am interested in is how I am interpreting it (it's academic really), so if my answer seems right in that context, then I'll leave it at that
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:40
add a comment |
$begingroup$
I think the issue is how to interpret the question. You are interpreting it as "what's the probability that the first two jelly beans are both white" but I'd say that was not what was intended. Rather, I think the idea was that someone draws them all out one at a time and asks for the probability that at some point two white ones are drawn consecutively.
$endgroup$
– lulu
Dec 3 '18 at 23:37
$begingroup$
You really should ask "the person much smarter than you" about how they interpret this question. I find your solution to be correct according to the wording of your question.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:58
$begingroup$
They only said the answer involves combinatorics. In any case, the question I am interested in is how I am interpreting it (it's academic really), so if my answer seems right in that context, then I'll leave it at that
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:40
$begingroup$
I think the issue is how to interpret the question. You are interpreting it as "what's the probability that the first two jelly beans are both white" but I'd say that was not what was intended. Rather, I think the idea was that someone draws them all out one at a time and asks for the probability that at some point two white ones are drawn consecutively.
$endgroup$
– lulu
Dec 3 '18 at 23:37
$begingroup$
I think the issue is how to interpret the question. You are interpreting it as "what's the probability that the first two jelly beans are both white" but I'd say that was not what was intended. Rather, I think the idea was that someone draws them all out one at a time and asks for the probability that at some point two white ones are drawn consecutively.
$endgroup$
– lulu
Dec 3 '18 at 23:37
$begingroup$
You really should ask "the person much smarter than you" about how they interpret this question. I find your solution to be correct according to the wording of your question.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:58
$begingroup$
You really should ask "the person much smarter than you" about how they interpret this question. I find your solution to be correct according to the wording of your question.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:58
$begingroup$
They only said the answer involves combinatorics. In any case, the question I am interested in is how I am interpreting it (it's academic really), so if my answer seems right in that context, then I'll leave it at that
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:40
$begingroup$
They only said the answer involves combinatorics. In any case, the question I am interested in is how I am interpreting it (it's academic really), so if my answer seems right in that context, then I'll leave it at that
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your answer looks fine to me. Did the other person give a reason as to why this would not be correct?
To be formal, let $W_i$ be the event that the $i^{th}$ jelly bean I select is white, assuming each jelly bean is eaten after selection. We want to find $Pr[W_1 cap W_2]$. But this can be expanded as $Pr[W_1 cap W_2] = Pr[W_1] cdot Pr[W_2 mid W_1]$. Like you noted above, $Pr[W_1] = frac{22}{60}$ and $Pr[W_2 mid W_1] = frac{21}{59}$, assuming every bean in the jar is equally likely to be chosen. This results in the answer you give above.
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
$endgroup$
$begingroup$
So, we both agreed that Pr[W1] = 22/60. We disagreed on Pr[W2 | W1] ... I said 21/59 and he said he doesn't remember but he knows it's not that. I'm not going to press him on it, I'm just curious if he is overthinking it and needlessly confusing me.
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:47
1
$begingroup$
Nope; if you are choosing without replacement, then your answer is indeed correct.
$endgroup$
– platty
Dec 6 '18 at 19:54
add a comment |
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1 Answer
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$begingroup$
Your answer looks fine to me. Did the other person give a reason as to why this would not be correct?
To be formal, let $W_i$ be the event that the $i^{th}$ jelly bean I select is white, assuming each jelly bean is eaten after selection. We want to find $Pr[W_1 cap W_2]$. But this can be expanded as $Pr[W_1 cap W_2] = Pr[W_1] cdot Pr[W_2 mid W_1]$. Like you noted above, $Pr[W_1] = frac{22}{60}$ and $Pr[W_2 mid W_1] = frac{21}{59}$, assuming every bean in the jar is equally likely to be chosen. This results in the answer you give above.
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
$endgroup$
$begingroup$
So, we both agreed that Pr[W1] = 22/60. We disagreed on Pr[W2 | W1] ... I said 21/59 and he said he doesn't remember but he knows it's not that. I'm not going to press him on it, I'm just curious if he is overthinking it and needlessly confusing me.
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:47
1
$begingroup$
Nope; if you are choosing without replacement, then your answer is indeed correct.
$endgroup$
– platty
Dec 6 '18 at 19:54
add a comment |
$begingroup$
Your answer looks fine to me. Did the other person give a reason as to why this would not be correct?
To be formal, let $W_i$ be the event that the $i^{th}$ jelly bean I select is white, assuming each jelly bean is eaten after selection. We want to find $Pr[W_1 cap W_2]$. But this can be expanded as $Pr[W_1 cap W_2] = Pr[W_1] cdot Pr[W_2 mid W_1]$. Like you noted above, $Pr[W_1] = frac{22}{60}$ and $Pr[W_2 mid W_1] = frac{21}{59}$, assuming every bean in the jar is equally likely to be chosen. This results in the answer you give above.
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
$endgroup$
$begingroup$
So, we both agreed that Pr[W1] = 22/60. We disagreed on Pr[W2 | W1] ... I said 21/59 and he said he doesn't remember but he knows it's not that. I'm not going to press him on it, I'm just curious if he is overthinking it and needlessly confusing me.
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:47
1
$begingroup$
Nope; if you are choosing without replacement, then your answer is indeed correct.
$endgroup$
– platty
Dec 6 '18 at 19:54
add a comment |
$begingroup$
Your answer looks fine to me. Did the other person give a reason as to why this would not be correct?
To be formal, let $W_i$ be the event that the $i^{th}$ jelly bean I select is white, assuming each jelly bean is eaten after selection. We want to find $Pr[W_1 cap W_2]$. But this can be expanded as $Pr[W_1 cap W_2] = Pr[W_1] cdot Pr[W_2 mid W_1]$. Like you noted above, $Pr[W_1] = frac{22}{60}$ and $Pr[W_2 mid W_1] = frac{21}{59}$, assuming every bean in the jar is equally likely to be chosen. This results in the answer you give above.
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
$endgroup$
Your answer looks fine to me. Did the other person give a reason as to why this would not be correct?
To be formal, let $W_i$ be the event that the $i^{th}$ jelly bean I select is white, assuming each jelly bean is eaten after selection. We want to find $Pr[W_1 cap W_2]$. But this can be expanded as $Pr[W_1 cap W_2] = Pr[W_1] cdot Pr[W_2 mid W_1]$. Like you noted above, $Pr[W_1] = frac{22}{60}$ and $Pr[W_2 mid W_1] = frac{21}{59}$, assuming every bean in the jar is equally likely to be chosen. This results in the answer you give above.
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
answered Dec 3 '18 at 23:27
plattyplatty
3,370320
3,370320
$begingroup$
So, we both agreed that Pr[W1] = 22/60. We disagreed on Pr[W2 | W1] ... I said 21/59 and he said he doesn't remember but he knows it's not that. I'm not going to press him on it, I'm just curious if he is overthinking it and needlessly confusing me.
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:47
1
$begingroup$
Nope; if you are choosing without replacement, then your answer is indeed correct.
$endgroup$
– platty
Dec 6 '18 at 19:54
add a comment |
$begingroup$
So, we both agreed that Pr[W1] = 22/60. We disagreed on Pr[W2 | W1] ... I said 21/59 and he said he doesn't remember but he knows it's not that. I'm not going to press him on it, I'm just curious if he is overthinking it and needlessly confusing me.
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:47
1
$begingroup$
Nope; if you are choosing without replacement, then your answer is indeed correct.
$endgroup$
– platty
Dec 6 '18 at 19:54
$begingroup$
So, we both agreed that Pr[W1] = 22/60. We disagreed on Pr[W2 | W1] ... I said 21/59 and he said he doesn't remember but he knows it's not that. I'm not going to press him on it, I'm just curious if he is overthinking it and needlessly confusing me.
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:47
$begingroup$
So, we both agreed that Pr[W1] = 22/60. We disagreed on Pr[W2 | W1] ... I said 21/59 and he said he doesn't remember but he knows it's not that. I'm not going to press him on it, I'm just curious if he is overthinking it and needlessly confusing me.
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:47
1
1
$begingroup$
Nope; if you are choosing without replacement, then your answer is indeed correct.
$endgroup$
– platty
Dec 6 '18 at 19:54
$begingroup$
Nope; if you are choosing without replacement, then your answer is indeed correct.
$endgroup$
– platty
Dec 6 '18 at 19:54
add a comment |
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$begingroup$
I think the issue is how to interpret the question. You are interpreting it as "what's the probability that the first two jelly beans are both white" but I'd say that was not what was intended. Rather, I think the idea was that someone draws them all out one at a time and asks for the probability that at some point two white ones are drawn consecutively.
$endgroup$
– lulu
Dec 3 '18 at 23:37
$begingroup$
You really should ask "the person much smarter than you" about how they interpret this question. I find your solution to be correct according to the wording of your question.
$endgroup$
– EvanHehehe
Dec 4 '18 at 0:58
$begingroup$
They only said the answer involves combinatorics. In any case, the question I am interested in is how I am interpreting it (it's academic really), so if my answer seems right in that context, then I'll leave it at that
$endgroup$
– Elijah Rockers
Dec 6 '18 at 19:40