## 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 ?

1. dan815

=]

2. anonymous

P(at least 2 boys|1boy)

3. mathmath333

\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

how to solve this P(at least 2 boys|1boy) ?

5. dan815

you can click edit and put it in the blue part

6. mathmath333

thnks for suggestion

7. anonymous

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

is this like bayes therem

9. anonymous

no just conditional probability Bayes is a little different

10. mathmath333

P(at least 2 boys)/P(at least one boy)=(1/3+1/3)/(3/3)

11. anonymous

not quite... P(At least 2 boys) = P(2 boys) + P(3 boys) P(at least 1 boy) = 1-P(no boys)

12. mathmath333

=(1/3+1/3)/(1-0/3) =?

13. anonymous

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

15. dan815

|dw:1444505252205:dw|

16. dan815

the area of the smallest circle divded by the bigger circle

17. anonymous

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

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

P(at least 2 boys)/P(at least one boy)=(1/2)/(1/8)=1/4 , is it

20. dan815

|dw:1444505457548:dw|

21. dan815

|dw:1444505579498:dw|

22. mathmath333

but the answer given inn book is $$\large \dfrac34$$

23. mathmath333

i m very confused

24. dan815

i prolly messed up somewhere, trying a slightly weird method

25. dan815

pgpilots way is more intuitive

26. amistre64

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

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

oh amistres way makes sense xD

29. dan815

however im not too sure if thats all there is to it, like in his scneario the first one is always b

30. mathmath333

i m dipresed

31. dan815

lmao

32. dan815

why

33. mathmath333

34. dan815

we can see :)

35. dan815

with a program

36. dan815

hey its 4/7 ya

37. dan815

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

oh ok 4/7

39. mathmath333

i hope ur program is correct

40. freckles

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

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

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

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

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

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

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

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

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

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

50. freckles

what is the difference between bbg and bgb

51. freckles

both of those say 2 boys and one girl

52. amistre64

the sample space is preserved, in my mind

53. ybarrap

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

i tend to do card probabilities with P instead of C as well ... dont know why

55. freckles

how do we know age matters?

56. dan815

thats not how it works i thnk

57. dan815

u are only considering b first then

58. dan815

when u consider b atleast in all places and then see the other possibilties , the bbb case becomes repeated in them

59. dan815

so its not exactly 3/4 its actually a bit less

60. freckles

hey P(B_2)=3/7 right @ybarrap if you go with your new list you have

61. dan815

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

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

you can manually check its usually less than 3/4 and around 4/7

64. freckles

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

this is what @amistre64 was saying earlier I believe to far up to scroll

66. freckles

I think I agree with @amistre64 now

67. dan815

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

69. freckles

70. dan815

B doesnt ahve to come first though

71. dan815

they just want B anywhere

72. freckles

Then how did you all those extra possibility

73. dan815

watcha mean

74. freckles

Like if B has to come first then you only have: the possibilities I just suggested

75. dan815

yeah that is true

76. dan815

but B doesnt haev to come first so we gotta consider everything

77. freckles

oh I thought you said B had to come first

78. dan815

"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

80. freckles

I think

81. dan815

i dunno lemme think lol

82. dan815

83. dan815

|dw:1444514572870:dw|

84. dan815

yeah i did consider the different permutations BGB BBG GBB

85. freckles

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

and you can omit the GGG from the list above

87. freckles

oops nope I'm confused again

88. freckles

because BGB is the same as BBG

89. freckles

err...

90. amistre64

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

i think since the kids are already borned ... that age is not relevant

92. freckles

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

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

kinddd offfff

95. amistre64

is this a correct interpretation? lol ... in my case i never know

96. dan815

lol kind of i mean whats stopping us from just looking at everything

97. dan815

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

thats kind of the way i approached it

99. dan815

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

i removed the ones where it was all 3 girls i mean

101. dan815

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
103. Empty

Do we agree that there is a higher probability of there being 2 boys than there is of being 3 boys?

104. dan815

yes

105. Empty

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.

106. dan815

right

107. ybarrap

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

108. anonymous

probabilities amistre gave are not all equal... if they were, then it would be 3/4.