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mathmath333

  • one year ago

A person has 3 children with at least one boy. Find the probablitiy of having atleast 2 boys among the children ?

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  1. dan815
    • one year ago
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    =]

  2. anonymous
    • one year ago
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    P(at least 2 boys|1boy)

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{A person has 3 children with at least one boy.} \hspace{.33em}\\~\\ & \normalsize \text{Find the probablitiy of having atleast 2 boys } \hspace{.33em}\\~\\ & \normalsize \text{among the children ?} \hspace{.33em}\\~\\ \end{align}}\)

  4. mathmath333
    • one year ago
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    how to solve this P(at least 2 boys|1boy) ?

  5. dan815
    • one year ago
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    you can click edit and put it in the blue part

  6. mathmath333
    • one year ago
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    thnks for suggestion

  7. anonymous
    • one year ago
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    P(at least 2 boys | at least 1boy) = P(at least 2 boys and at least 1 boy)/P(at least one boy) P(at least 2 boys)/P(at least one boy)

  8. mathmath333
    • one year ago
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    is this like bayes therem

  9. anonymous
    • one year ago
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    no just conditional probability Bayes is a little different

  10. mathmath333
    • one year ago
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    P(at least 2 boys)/P(at least one boy)=(1/3+1/3)/(3/3)

  11. anonymous
    • one year ago
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    not quite... P(At least 2 boys) = P(2 boys) + P(3 boys) P(at least 1 boy) = 1-P(no boys)

  12. mathmath333
    • one year ago
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    =(1/3+1/3)/(1-0/3) =?

  13. anonymous
    • one year ago
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    P(no boys) = P(all girls) how many children does the couple have? 3. => P(all girls) = P(GGG) = (1/2)(1/2)(1/2) = 1/8

  14. mathmath333
    • one year ago
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    final answer=(2/3)/(1/8)=1/12

  15. dan815
    • one year ago
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    |dw:1444505252205:dw|

  16. dan815
    • one year ago
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    the area of the smallest circle divded by the bigger circle

  17. anonymous
    • one year ago
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    no P(At least 2 boys) = P(2 boys) + P(3 boys) P(3 boys) = P(3 girls) = 1/8 P(2 boys) = 3C2 * (1/8) = 3/8 P(At least 2 boys) = 1/2

  18. dan815
    • one year ago
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    Prob of atleast 2 boys = prob of atleast 1 boy - prob of only 1 boy prob of @2boys/prob of @ 1 boy = (prob of atleast 1 boy - prob of only 1 boy)/prob of @ 1 boy = 1- prob_of_1_boy/prob of @ 1 boy = 1- prob of 1 boy/(1-prob of no boys)

  19. mathmath333
    • one year ago
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    P(at least 2 boys)/P(at least one boy)=(1/2)/(1/8)=1/4 , is it

  20. dan815
    • one year ago
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    |dw:1444505457548:dw|

  21. dan815
    • one year ago
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    |dw:1444505579498:dw|

  22. mathmath333
    • one year ago
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    but the answer given inn book is \(\large \dfrac34\)

  23. mathmath333
    • one year ago
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    i m very confused

  24. dan815
    • one year ago
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    i prolly messed up somewhere, trying a slightly weird method

  25. dan815
    • one year ago
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    pgpilots way is more intuitive

  26. amistre64
    • one year ago
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    b gg b gb b bg b bb 3 out of 4 in the sample space have at least 2 bs, given 1b

  27. dan815
    • one year ago
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    p @least 2boys =p(2boys)+p(3boys)= 3/8 + 1/8=4/8 p@1boy = p@2 boys + p(1boy) = 4/8+3/8 herefore 4/8/7/8 = 4/7

  28. dan815
    • one year ago
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    oh amistres way makes sense xD

  29. dan815
    • one year ago
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    however im not too sure if thats all there is to it, like in his scneario the first one is always b

  30. mathmath333
    • one year ago
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    i m dipresed

  31. dan815
    • one year ago
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    lmao

  32. dan815
    • one year ago
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    why

  33. mathmath333
    • one year ago
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    which answer is correct

  34. dan815
    • one year ago
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    we can see :)

  35. dan815
    • one year ago
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    with a program

  36. dan815
    • one year ago
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    hey its 4/7 ya

  37. dan815
    • one year ago
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    d=randi(2) a=[] t=1000 for i =1:t for j=1:3 d=randi(2); if d==1 a(i,j)=1; else a(i,j)=0; end end end a; hold=[] for i = 1:t if sum(a(i,:))==0; hold(i)=1; end end b=sum(hold) %number of no 1 hold=[] for i=1:t if sum(a(i,:))==1; hold(i)=1; end end c=sum(hold)% number of only 1 prob=(t-(b+c))/(t-b) OUTPUT prob = 0.5716 and 4/7 = 0.5714

  38. mathmath333
    • one year ago
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    oh ok 4/7

  39. mathmath333
    • one year ago
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    i hope ur program is correct

  40. freckles
    • one year ago
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    I was thinking this as first: BBB BGG BBG GGG are the only possibilities because the following list give repeats: GBG since this is equal to BGG GGB since this is still BGG GBB since this is equal to BBG BGB since this is equal to BBG So we have 4 possibilities for a group 3 children: BBB BGG BBG GGG Now we have to find the probabilities of least 2 boys given at least 1 boy this means we are looking for \[P(2 \text{ or } 3|1 \text{ or } 2 \text{ or } 3) \\ =\frac{P( [2 \text{ or } 3] \cap [1 \text{ or } 2 \text{ or } 3] )}{P(1 \text{ or } 2 \text{ or } 3)}=\frac{P(2 \text{ or } 3)}{P(1 \text{ or } 2 \text{ or } 3)} =\frac{\frac{2}{4}}{\frac{3}{4}}=\frac{2}{3}\] .. but I'm still thinking...

  41. freckles
    • one year ago
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    however if I read that one thing as we have 3+1 kids where that 1 kid is boy and the other children or whatevers.. then following that one route from above we do get 3/4

  42. freckles
    • one year ago
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    like is it 3 kids +1 boy that is given or is it saying 3 kids where 1 of those 3 is a boy ?

  43. ybarrap
    • one year ago
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    If you don't consider order there are four distinct possibilities; BBB;GGG;GGB;BBG But we must consider order and so there are really eight possibilities: BBB;BBG;BGB;GBB;GGB;GBG;BGG;GGG Let \(B_i\) represent the event that there are \(i\) boys born out of 3. We are then looking for \(P(B_2)+P(B_3)\), the probability of 2 or more boys among the children: $$ P(B_2)=\cfrac{2}{8}\\ P(B_3)=\cfrac{1}{8}\\ P(B_2)+P(B_3)=\cfrac{3}{8} $$

  44. freckles
    • one year ago
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    BBBB BBBG BBGG BGGG GGGG this would be: \[\frac{P(2 \text{ or } 3 \text{ or } 4)}{P(1 \text{ or } 2 \text{ or } 3 \text{ or } 4)}=\frac{3}{4}\]

  45. freckles
    • one year ago
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    BBB;BBG;BGB;GBB;GGB;GBG;BGG;GGG aren't some of these possibilities equal? for example BBG=BGB=GBB or GGB=GBG=BGG

  46. amistre64
    • one year ago
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    A person has 3 children with at least one boy. b gg or b bg or b gb or b bb ggg is not even in the sample space

  47. freckles
    • one year ago
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    bbg=bgb I listed all the possibilities for 3 kids notice the answer didn't even care about the possibility that is GGG

  48. amistre64
    • one year ago
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    lets assume we are wanting to kill socrates :) we grab 3 of stones from a jar of black stones and white stones ... we show one stone, it is black. what is the probability that we have at least 2 black stones in our grips?

  49. freckles
    • one year ago
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    when you guys list the 8 possibilities does that mean we are concerned with which one popped out first, second, and third...

  50. freckles
    • one year ago
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    what is the difference between bbg and bgb

  51. freckles
    • one year ago
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    both of those say 2 boys and one girl

  52. amistre64
    • one year ago
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    the sample space is preserved, in my mind

  53. ybarrap
    • one year ago
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    That's right! Given one boy was born, the sample space should be: BBB;BBG;BGB;GBB;GGB;GBG;BGG $$ P(B_2)=\cfrac{2}{7}\\ P(B_3)=\cfrac{1}{7}\\ P(B_2)+P(B_3)=\cfrac{3}{7} $$ Order is important in this sample space because there are more than one way that a group of 3 children can have two girls or two boys, but only one way to have all boys.The relative age of the child is relevant.

  54. amistre64
    • one year ago
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    i tend to do card probabilities with P instead of C as well ... dont know why

  55. freckles
    • one year ago
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    how do we know age matters?

  56. dan815
    • one year ago
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    thats not how it works i thnk

  57. dan815
    • one year ago
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    u are only considering b first then

  58. dan815
    • one year ago
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    when u consider b atleast in all places and then see the other possibilties , the bbb case becomes repeated in them

  59. dan815
    • one year ago
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    so its not exactly 3/4 its actually a bit less

  60. freckles
    • one year ago
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    hey P(B_2)=3/7 right @ybarrap if you go with your new list you have

  61. dan815
    • one year ago
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    here's some trial space a = 0 0 0 1 0 1 1 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 0 0 1 0 a = 0 1 1 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 0 1 1 1 1 1 1

  62. dan815
    • one year ago
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    a = 0 0 0 0 1 0 1 0 0 1 1 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 a = 0 0 1 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 a = 1 1 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 0 0 1 1

  63. dan815
    • one year ago
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    you can manually check its usually less than 3/4 and around 4/7

  64. freckles
    • one year ago
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    ok if B has to come first then we have this list: B|BB B|GG B|BG B|GB There are 4 possibilities P(2)+P(3)=2/4+1/4=3/4

  65. freckles
    • one year ago
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    this is what @amistre64 was saying earlier I believe to far up to scroll

  66. freckles
    • one year ago
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    I think I agree with @amistre64 now

  67. dan815
    • one year ago
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    p @least 2boys =p(2boys)+p(3boys)= 3/8 + 1/8=4/8 p@least 1boy = p@2 boys + p(1boy) = 4/8+3/8 p(@least2boys|@1boy) = 4/8/7/8 = 4/7

  68. dan815
    • one year ago
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    the book answer is wrong

  69. freckles
    • one year ago
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    but isn't your answer suggesting B doesn't have to come first?

  70. dan815
    • one year ago
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    B doesnt ahve to come first though

  71. dan815
    • one year ago
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    they just want B anywhere

  72. freckles
    • one year ago
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    Then how did you all those extra possibility

  73. dan815
    • one year ago
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    watcha mean

  74. freckles
    • one year ago
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    Like if B has to come first then you only have: the possibilities I just suggested

  75. dan815
    • one year ago
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    yeah that is true

  76. dan815
    • one year ago
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    but B doesnt haev to come first so we gotta consider everything

  77. freckles
    • one year ago
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    oh I thought you said B had to come first

  78. dan815
    • one year ago
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    "A person has 3 children with at least one boy.Find the probablitiy of having atleast 2 boys among the children ?" the question just wants 1 boy anywhere

  79. freckles
    • one year ago
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    the answer suggest the book doesn't care about age your answer suggest the book cares about age

  80. freckles
    • one year ago
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    I think

  81. dan815
    • one year ago
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    i dunno lemme think lol

  82. dan815
    • one year ago
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    i was just think about this picture

  83. dan815
    • one year ago
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    |dw:1444514572870:dw|

  84. dan815
    • one year ago
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    yeah i did consider the different permutations BGB BBG GBB

  85. freckles
    • one year ago
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    BBB BGG BGB BBG GGG GBB GBG GGB is the list I think you are using and for example BGG is the same as GGB; like both say two girls and one boy But we included them both because BGG is saying the boy came first then that one girl than that other girl --- BBB BGB BBG BGG this one doesn't care about age at all

  86. freckles
    • one year ago
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    and you can omit the GGG from the list above

  87. freckles
    • one year ago
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    oops nope I'm confused again

  88. freckles
    • one year ago
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    because BGB is the same as BBG

  89. freckles
    • one year ago
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    err...

  90. amistre64
    • one year ago
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    there are 3 kids standing behind a curtain .. lol the first one steps out ... its a boy!! what is the probability that there were at least 2 boys standing behind the curtain? i just listed the ways that the other 2 kids could come forth ....

  91. amistre64
    • one year ago
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    i think since the kids are already borned ... that age is not relevant

  92. freckles
    • one year ago
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    so @amistre64 you are saying the first kid has to by a boy but the order doesn't matter for the other two right?

  93. amistre64
    • one year ago
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    given that we know one if the kids is a boy .... so its assumed that we have this prior knowledge of the sample space yes.

  94. dan815
    • one year ago
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    kinddd offfff

  95. amistre64
    • one year ago
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    is this a correct interpretation? lol ... in my case i never know

  96. dan815
    • one year ago
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    lol kind of i mean whats stopping us from just looking at everything

  97. dan815
    • one year ago
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    like forget they even say its atleast 1 boy, cant we just solve it like its 1/2 chance for each and see every single possilbility and then check for ourselves

  98. dan815
    • one year ago
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    thats kind of the way i approached it

  99. dan815
    • one year ago
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    and to check i made a code that simply spat out 3 kids, boys or girls, and i removed the all Gs from them, and i looked at the ratio @least 2 bots/@atleast 1 boy

  100. dan815
    • one year ago
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    i removed the ones where it was all 3 girls i mean

  101. dan815
    • one year ago
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    d=randi(2) a=[] t=10000, %The number of sets of 3 children %Creats the Random Trial space for i =1:t for j=1:3 d=randi(2); if d==1 a(i,j)=1; else a(i,j)=0; end end end a; %Find the number of cases where all 3 children were girls hold=[] for i = 1:t if sum(a(i,:))==0; hold(i)=1; end end b=sum(hold) %#of all G's set %Finds the number of case where there is only 1 boy hold=[] for i=1:t if sum(a(i,:))==1; hold(i)=1; end end c=sum(hold) % #of only 1 boy sets %ANSWER prob=(t-(b+c))/(t-b) %atleast2boys/atleast1boy %atleast 2 boys = #of Total sets- #of all G's set-#of only 1 boy sets %atleast 1 boy = #of Total sets - #of all G's set

  102. Empty
    • one year ago
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    Do we agree that there is a higher probability of there being 2 boys than there is of being 3 boys?

  103. dan815
    • one year ago
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    yes

  104. Empty
    • one year ago
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    If you agree then you know that it can't possibly be 3/4 since this is simply considering only: GGG, BGG, BBG, BBB but we know BBG has a higher probability of happening than BBB, so 3/4 is wrong. This isn't a combinatorics question.

  105. dan815
    • one year ago
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    right

  106. ybarrap
    • one year ago
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    Thanks @freckles! This probability should be: $$ P(B_2)=\cfrac{3}{7}\\ P(B_3)=\cfrac{1}{7}\\ P(B_2)+P(B_3)=\cfrac{4}{7} $$ The reason is that 2 boys is more likely than 3 boys is the same reason that 2 heads is more likely than 3 heads. For 2 heads you can have hht, hth or thh, while for three heads you only have only hhh as a possibility. So the age/order of the boys matters since the order of heads matters. The sample spaces for 2 or more tails/boys given at least one tail/boy is: BBB;BBG;BGB;GBB;GGB;GBG;BGG TTT;TTH;THT;HTT;HHT;HTH;THH

  107. anonymous
    • one year ago
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    probabilities amistre gave are not all equal... if they were, then it would be 3/4.

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