Combinatorics cookie question – what’s wrong with my method?












2












$begingroup$


I’d like to ask what’s wrong with my method for this supposedly easy question:




There are 4 fresh cookies and 2 spoilt ones. Two children grab 2 cookies each at random. What is the probability that both children have one fresh and one spoilt cookie each?




The approach I have used is a simple one. The total number of ways to distribute the cookies is ${6choose 2}times{4choose 2} = 90$, and the number of ways to meet the condition that both children have one spoilt and one fresh cookie each is $({4choose 1}times{2choose 1})times({3choose 1}times 1) = 24$. The reason for the former is that the first child has 6 cookies to choose from, and the second child has 4; while for the latter, each child must choose 1 cookie from each group. The probability is therefore $frac{24}{90} = frac{4}{15}$.



The correct answer, however, $frac{2}{3}$. The logic is that either kid gets the first spoilt cookie, and the probability he picks the other spoilt cookie is $frac{1}{3}$, and hence the probability of meeting the condition is the complement. I understand this logic, however I can’t find the flaws in mine – where did I go wrong? Thank you!






probability combinatorics







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    The supposedly correct method seems flawed. There is no need for either kid to get a spoiled cookie, they could both take two fresh ones. And even if one kid first selects a bad cookie, the probability is then $frac 15$ that the same kid then selects the second bad one
    $endgroup$
    – lulu
    Dec 22 '18 at 16:40












  • $begingroup$
    Note: you keep referring to "one spoilt and one mouldy" cookie but surely the mouldy ones coincide with the spoilt ones. I assume you meant to say "one fresh and one spoilt" one.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:44










  • $begingroup$
    Given the correction I pointed out, I agree with your answer of $frac 4{15}$. The supposedly correct answer of $frac 23$ is absurdly high. There is, after all, only a $frac 8{15}$ probability that the first child will select one fresh and one bad cookie and, conditioned on that, a smaller chance for the second child to do the same.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:46








  • 1




    $begingroup$
    Your “correct answer” is the conditional probability that each child gets one spoiled cookie, given that the two unchosen cookies are fresh. The probability that the two unchosen cookies are fresh is ${4over6}times{3over5}$, so the answer to the question is ${4over6}times{3over5}times{2over3}={4over{15}}$.
    $endgroup$
    – Steve Kass
    Dec 22 '18 at 20:18






  • 2




    $begingroup$
    Hmm, The last comment of farruhota's answer implies the question is if there were four cookies total. then that logic works. All cookies are grabbed so the Bad Cookie A was grabbed. There are 3 other cookies so the prob the kid we grabbed bad cookie A also grabbed Bad Coookie B is 1/3. Or your way there are ${4choose 2}{2choose 2}=6$ ways cookies can be had. And $({2choose 1}{2choose 1})times({1choose 1}{1choose 1})=4$ for each to choose one bad and one good. so prob is $frac 46=frac 23$
    $endgroup$
    – fleablood
    Dec 22 '18 at 20:23


















2












$begingroup$


I’d like to ask what’s wrong with my method for this supposedly easy question:




There are 4 fresh cookies and 2 spoilt ones. Two children grab 2 cookies each at random. What is the probability that both children have one fresh and one spoilt cookie each?




The approach I have used is a simple one. The total number of ways to distribute the cookies is ${6choose 2}times{4choose 2} = 90$, and the number of ways to meet the condition that both children have one spoilt and one fresh cookie each is $({4choose 1}times{2choose 1})times({3choose 1}times 1) = 24$. The reason for the former is that the first child has 6 cookies to choose from, and the second child has 4; while for the latter, each child must choose 1 cookie from each group. The probability is therefore $frac{24}{90} = frac{4}{15}$.



The correct answer, however, $frac{2}{3}$. The logic is that either kid gets the first spoilt cookie, and the probability he picks the other spoilt cookie is $frac{1}{3}$, and hence the probability of meeting the condition is the complement. I understand this logic, however I can’t find the flaws in mine – where did I go wrong? Thank you!






probability combinatorics







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    The supposedly correct method seems flawed. There is no need for either kid to get a spoiled cookie, they could both take two fresh ones. And even if one kid first selects a bad cookie, the probability is then $frac 15$ that the same kid then selects the second bad one
    $endgroup$
    – lulu
    Dec 22 '18 at 16:40












  • $begingroup$
    Note: you keep referring to "one spoilt and one mouldy" cookie but surely the mouldy ones coincide with the spoilt ones. I assume you meant to say "one fresh and one spoilt" one.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:44










  • $begingroup$
    Given the correction I pointed out, I agree with your answer of $frac 4{15}$. The supposedly correct answer of $frac 23$ is absurdly high. There is, after all, only a $frac 8{15}$ probability that the first child will select one fresh and one bad cookie and, conditioned on that, a smaller chance for the second child to do the same.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:46








  • 1




    $begingroup$
    Your “correct answer” is the conditional probability that each child gets one spoiled cookie, given that the two unchosen cookies are fresh. The probability that the two unchosen cookies are fresh is ${4over6}times{3over5}$, so the answer to the question is ${4over6}times{3over5}times{2over3}={4over{15}}$.
    $endgroup$
    – Steve Kass
    Dec 22 '18 at 20:18






  • 2




    $begingroup$
    Hmm, The last comment of farruhota's answer implies the question is if there were four cookies total. then that logic works. All cookies are grabbed so the Bad Cookie A was grabbed. There are 3 other cookies so the prob the kid we grabbed bad cookie A also grabbed Bad Coookie B is 1/3. Or your way there are ${4choose 2}{2choose 2}=6$ ways cookies can be had. And $({2choose 1}{2choose 1})times({1choose 1}{1choose 1})=4$ for each to choose one bad and one good. so prob is $frac 46=frac 23$
    $endgroup$
    – fleablood
    Dec 22 '18 at 20:23
















2












2








2





$begingroup$


I’d like to ask what’s wrong with my method for this supposedly easy question:




There are 4 fresh cookies and 2 spoilt ones. Two children grab 2 cookies each at random. What is the probability that both children have one fresh and one spoilt cookie each?




The approach I have used is a simple one. The total number of ways to distribute the cookies is ${6choose 2}times{4choose 2} = 90$, and the number of ways to meet the condition that both children have one spoilt and one fresh cookie each is $({4choose 1}times{2choose 1})times({3choose 1}times 1) = 24$. The reason for the former is that the first child has 6 cookies to choose from, and the second child has 4; while for the latter, each child must choose 1 cookie from each group. The probability is therefore $frac{24}{90} = frac{4}{15}$.



The correct answer, however, $frac{2}{3}$. The logic is that either kid gets the first spoilt cookie, and the probability he picks the other spoilt cookie is $frac{1}{3}$, and hence the probability of meeting the condition is the complement. I understand this logic, however I can’t find the flaws in mine – where did I go wrong? Thank you!






probability combinatorics







share|cite|improve this question











$endgroup$




I’d like to ask what’s wrong with my method for this supposedly easy question:




There are 4 fresh cookies and 2 spoilt ones. Two children grab 2 cookies each at random. What is the probability that both children have one fresh and one spoilt cookie each?




The approach I have used is a simple one. The total number of ways to distribute the cookies is ${6choose 2}times{4choose 2} = 90$, and the number of ways to meet the condition that both children have one spoilt and one fresh cookie each is $({4choose 1}times{2choose 1})times({3choose 1}times 1) = 24$. The reason for the former is that the first child has 6 cookies to choose from, and the second child has 4; while for the latter, each child must choose 1 cookie from each group. The probability is therefore $frac{24}{90} = frac{4}{15}$.



The correct answer, however, $frac{2}{3}$. The logic is that either kid gets the first spoilt cookie, and the probability he picks the other spoilt cookie is $frac{1}{3}$, and hence the probability of meeting the condition is the complement. I understand this logic, however I can’t find the flaws in mine – where did I go wrong? Thank you!






probability combinatorics




probability combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 17:05







user107224

















asked Dec 22 '18 at 16:36









user107224user107224

432314




432314








  • 6




    $begingroup$
    The supposedly correct method seems flawed. There is no need for either kid to get a spoiled cookie, they could both take two fresh ones. And even if one kid first selects a bad cookie, the probability is then $frac 15$ that the same kid then selects the second bad one
    $endgroup$
    – lulu
    Dec 22 '18 at 16:40












  • $begingroup$
    Note: you keep referring to "one spoilt and one mouldy" cookie but surely the mouldy ones coincide with the spoilt ones. I assume you meant to say "one fresh and one spoilt" one.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:44










  • $begingroup$
    Given the correction I pointed out, I agree with your answer of $frac 4{15}$. The supposedly correct answer of $frac 23$ is absurdly high. There is, after all, only a $frac 8{15}$ probability that the first child will select one fresh and one bad cookie and, conditioned on that, a smaller chance for the second child to do the same.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:46








  • 1




    $begingroup$
    Your “correct answer” is the conditional probability that each child gets one spoiled cookie, given that the two unchosen cookies are fresh. The probability that the two unchosen cookies are fresh is ${4over6}times{3over5}$, so the answer to the question is ${4over6}times{3over5}times{2over3}={4over{15}}$.
    $endgroup$
    – Steve Kass
    Dec 22 '18 at 20:18






  • 2




    $begingroup$
    Hmm, The last comment of farruhota's answer implies the question is if there were four cookies total. then that logic works. All cookies are grabbed so the Bad Cookie A was grabbed. There are 3 other cookies so the prob the kid we grabbed bad cookie A also grabbed Bad Coookie B is 1/3. Or your way there are ${4choose 2}{2choose 2}=6$ ways cookies can be had. And $({2choose 1}{2choose 1})times({1choose 1}{1choose 1})=4$ for each to choose one bad and one good. so prob is $frac 46=frac 23$
    $endgroup$
    – fleablood
    Dec 22 '18 at 20:23
















  • 6




    $begingroup$
    The supposedly correct method seems flawed. There is no need for either kid to get a spoiled cookie, they could both take two fresh ones. And even if one kid first selects a bad cookie, the probability is then $frac 15$ that the same kid then selects the second bad one
    $endgroup$
    – lulu
    Dec 22 '18 at 16:40












  • $begingroup$
    Note: you keep referring to "one spoilt and one mouldy" cookie but surely the mouldy ones coincide with the spoilt ones. I assume you meant to say "one fresh and one spoilt" one.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:44










  • $begingroup$
    Given the correction I pointed out, I agree with your answer of $frac 4{15}$. The supposedly correct answer of $frac 23$ is absurdly high. There is, after all, only a $frac 8{15}$ probability that the first child will select one fresh and one bad cookie and, conditioned on that, a smaller chance for the second child to do the same.
    $endgroup$
    – lulu
    Dec 22 '18 at 16:46








  • 1




    $begingroup$
    Your “correct answer” is the conditional probability that each child gets one spoiled cookie, given that the two unchosen cookies are fresh. The probability that the two unchosen cookies are fresh is ${4over6}times{3over5}$, so the answer to the question is ${4over6}times{3over5}times{2over3}={4over{15}}$.
    $endgroup$
    – Steve Kass
    Dec 22 '18 at 20:18






  • 2




    $begingroup$
    Hmm, The last comment of farruhota's answer implies the question is if there were four cookies total. then that logic works. All cookies are grabbed so the Bad Cookie A was grabbed. There are 3 other cookies so the prob the kid we grabbed bad cookie A also grabbed Bad Coookie B is 1/3. Or your way there are ${4choose 2}{2choose 2}=6$ ways cookies can be had. And $({2choose 1}{2choose 1})times({1choose 1}{1choose 1})=4$ for each to choose one bad and one good. so prob is $frac 46=frac 23$
    $endgroup$
    – fleablood
    Dec 22 '18 at 20:23










6




6




$begingroup$
The supposedly correct method seems flawed. There is no need for either kid to get a spoiled cookie, they could both take two fresh ones. And even if one kid first selects a bad cookie, the probability is then $frac 15$ that the same kid then selects the second bad one
$endgroup$
– lulu
Dec 22 '18 at 16:40






$begingroup$
The supposedly correct method seems flawed. There is no need for either kid to get a spoiled cookie, they could both take two fresh ones. And even if one kid first selects a bad cookie, the probability is then $frac 15$ that the same kid then selects the second bad one
$endgroup$
– lulu
Dec 22 '18 at 16:40














$begingroup$
Note: you keep referring to "one spoilt and one mouldy" cookie but surely the mouldy ones coincide with the spoilt ones. I assume you meant to say "one fresh and one spoilt" one.
$endgroup$
– lulu
Dec 22 '18 at 16:44




$begingroup$
Note: you keep referring to "one spoilt and one mouldy" cookie but surely the mouldy ones coincide with the spoilt ones. I assume you meant to say "one fresh and one spoilt" one.
$endgroup$
– lulu
Dec 22 '18 at 16:44












$begingroup$
Given the correction I pointed out, I agree with your answer of $frac 4{15}$. The supposedly correct answer of $frac 23$ is absurdly high. There is, after all, only a $frac 8{15}$ probability that the first child will select one fresh and one bad cookie and, conditioned on that, a smaller chance for the second child to do the same.
$endgroup$
– lulu
Dec 22 '18 at 16:46






$begingroup$
Given the correction I pointed out, I agree with your answer of $frac 4{15}$. The supposedly correct answer of $frac 23$ is absurdly high. There is, after all, only a $frac 8{15}$ probability that the first child will select one fresh and one bad cookie and, conditioned on that, a smaller chance for the second child to do the same.
$endgroup$
– lulu
Dec 22 '18 at 16:46






1




1




$begingroup$
Your “correct answer” is the conditional probability that each child gets one spoiled cookie, given that the two unchosen cookies are fresh. The probability that the two unchosen cookies are fresh is ${4over6}times{3over5}$, so the answer to the question is ${4over6}times{3over5}times{2over3}={4over{15}}$.
$endgroup$
– Steve Kass
Dec 22 '18 at 20:18




$begingroup$
Your “correct answer” is the conditional probability that each child gets one spoiled cookie, given that the two unchosen cookies are fresh. The probability that the two unchosen cookies are fresh is ${4over6}times{3over5}$, so the answer to the question is ${4over6}times{3over5}times{2over3}={4over{15}}$.
$endgroup$
– Steve Kass
Dec 22 '18 at 20:18




2




2




$begingroup$
Hmm, The last comment of farruhota's answer implies the question is if there were four cookies total. then that logic works. All cookies are grabbed so the Bad Cookie A was grabbed. There are 3 other cookies so the prob the kid we grabbed bad cookie A also grabbed Bad Coookie B is 1/3. Or your way there are ${4choose 2}{2choose 2}=6$ ways cookies can be had. And $({2choose 1}{2choose 1})times({1choose 1}{1choose 1})=4$ for each to choose one bad and one good. so prob is $frac 46=frac 23$
$endgroup$
– fleablood
Dec 22 '18 at 20:23






$begingroup$
Hmm, The last comment of farruhota's answer implies the question is if there were four cookies total. then that logic works. All cookies are grabbed so the Bad Cookie A was grabbed. There are 3 other cookies so the prob the kid we grabbed bad cookie A also grabbed Bad Coookie B is 1/3. Or your way there are ${4choose 2}{2choose 2}=6$ ways cookies can be had. And $({2choose 1}{2choose 1})times({1choose 1}{1choose 1})=4$ for each to choose one bad and one good. so prob is $frac 46=frac 23$
$endgroup$
– fleablood
Dec 22 '18 at 20:23












1 Answer
1






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oldest

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$begingroup$

You can verify your answer with the probabilities:
$$P(F_1S_2F_3S_4)=frac 46cdot frac25cdot frac34 cdot frac 13=frac1 {15};\
P(F_1S_2S_3F_4)=frac 46cdot frac25cdot frac14 cdot frac 33=frac1 {15};\P(S_1F_2F_3S_4)=frac 26cdot frac45cdot frac34 cdot frac 13=frac1 {15};\P(S_1F_2S_3F_4)=frac 26cdot frac45cdot frac14 cdot frac 33=frac1 {15}.$$
Hence, the answer is $frac4{15}$.



I think the question must have implied there are 4 cookies overall and 2 of them are spoilt. Then the answer given works.






share|cite|improve this answer









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    $begingroup$

    You can verify your answer with the probabilities:
    $$P(F_1S_2F_3S_4)=frac 46cdot frac25cdot frac34 cdot frac 13=frac1 {15};\
    P(F_1S_2S_3F_4)=frac 46cdot frac25cdot frac14 cdot frac 33=frac1 {15};\P(S_1F_2F_3S_4)=frac 26cdot frac45cdot frac34 cdot frac 13=frac1 {15};\P(S_1F_2S_3F_4)=frac 26cdot frac45cdot frac14 cdot frac 33=frac1 {15}.$$
    Hence, the answer is $frac4{15}$.



    I think the question must have implied there are 4 cookies overall and 2 of them are spoilt. Then the answer given works.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You can verify your answer with the probabilities:
      $$P(F_1S_2F_3S_4)=frac 46cdot frac25cdot frac34 cdot frac 13=frac1 {15};\
      P(F_1S_2S_3F_4)=frac 46cdot frac25cdot frac14 cdot frac 33=frac1 {15};\P(S_1F_2F_3S_4)=frac 26cdot frac45cdot frac34 cdot frac 13=frac1 {15};\P(S_1F_2S_3F_4)=frac 26cdot frac45cdot frac14 cdot frac 33=frac1 {15}.$$
      Hence, the answer is $frac4{15}$.



      I think the question must have implied there are 4 cookies overall and 2 of them are spoilt. Then the answer given works.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You can verify your answer with the probabilities:
        $$P(F_1S_2F_3S_4)=frac 46cdot frac25cdot frac34 cdot frac 13=frac1 {15};\
        P(F_1S_2S_3F_4)=frac 46cdot frac25cdot frac14 cdot frac 33=frac1 {15};\P(S_1F_2F_3S_4)=frac 26cdot frac45cdot frac34 cdot frac 13=frac1 {15};\P(S_1F_2S_3F_4)=frac 26cdot frac45cdot frac14 cdot frac 33=frac1 {15}.$$
        Hence, the answer is $frac4{15}$.



        I think the question must have implied there are 4 cookies overall and 2 of them are spoilt. Then the answer given works.






        share|cite|improve this answer









        $endgroup$



        You can verify your answer with the probabilities:
        $$P(F_1S_2F_3S_4)=frac 46cdot frac25cdot frac34 cdot frac 13=frac1 {15};\
        P(F_1S_2S_3F_4)=frac 46cdot frac25cdot frac14 cdot frac 33=frac1 {15};\P(S_1F_2F_3S_4)=frac 26cdot frac45cdot frac34 cdot frac 13=frac1 {15};\P(S_1F_2S_3F_4)=frac 26cdot frac45cdot frac14 cdot frac 33=frac1 {15}.$$
        Hence, the answer is $frac4{15}$.



        I think the question must have implied there are 4 cookies overall and 2 of them are spoilt. Then the answer given works.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 20:07









        farruhotafarruhota

        20.2k2738




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            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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