Jelly bean choosing from a jar.












0












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 19:44







Elijah Rockers

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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