Statistics help please!

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Probability:

A couple has 2 kids and it is given that one of them is a girl. What is the probability that the other is a girl too? Use conditional probability.

(Please do not say 1/2 without solving it).

Thanks

Last edited: 05-Oct-10 11:01 PM

Arrogant

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The answer is 1/3

Alright, Let me try to Solve it.

So, you said Couple has 2 kids and one of them is girl. So, The combination of having 2 kids (Boy and Girl) would be :

Girl - Girl
Girl--Boy
Boy-Girl
Boy-Boy

Since one of them is already a Girl which leads us to elimination of Boy-Boy combination. Therefore, the probability that the other is a girl to is 1/3 .

dimag kharab

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If i were you i would have an Ultrasound instead of Using the conditional probability.

Stiffler

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Dimag, the question says the couple has two kids. Why would u do an ultrasound on kids who r already born?

In all practicality the answer should be 1/2 though I see where Arrogant is going.

YoTaBhayanaNiSom

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Stiffler:

This is Bayesian statistics. It may sound unintuitive to come up with 1/3 but it is the correct answer.

Last edited: 06-Oct-10 08:25 AM

dimag kharab

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Ahh! Got it!

In present world, 1/2 for either a boy or girl does not work.

The possibility of being a Trans, Gay or Bisexual should be considered too.

Damn! Statistics is getting difficult.

tregor

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arrogant, thanks

yotabhayenanisom, could u use the Baysean formulae to come up with the answer..?

thanks

hariyo

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dont take this:bb
take this sample::bg,gb,gg

wi=event that ith child is girl.

p(w2) = p(w2|w1)p(w1)+p(w2|not w1)p(not w1)
= {gg}{gb,gg}+{bg}{bg}
= 1/3*2/3 + 1/3*1/3
=1/3

YoTaBhayanaNiSom

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I thought this one is even more interesting and less intuitive.

A game-show climax draws nigh. A drum-roll sounds. The game show host leads you to a wall with three closed doors. Behind one of the doors is the automobile of your dreams, and behind each of the other two is a can of dog food. The three doors all have even chances of hiding the automobile. The host, a trustworthy person who knows precisely what is behind each of the three doors, explains how the game will work. First, you will choose a door without opening it, knowing that after you have done so, the host will open one of the two remaining doors to reveal a can of dog food. When this has been done, you will be given the opportunity to switch doors; you will win whatever is behind the door you choose at this stage of the game. Do you raise your chances of winning the automobile by switching doors?

tregor

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Actually I know the answer ...there is a mathematical explanation on why we should switch or keep...but I will let other people try first

tregor

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thanks Hariyo

hariyo

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This is famous game.

1> He doesn't switch
if he choose prize door
wi= win in ith choose
p(w1) = p(w2)=1/3

if he choose wrong door(no prize)
p(w1) = 0

2>He switch to the other remaining door than shown by host

if he was in prize door and he switch
p(w2|w1) = 0
if he was not in prize door ,host shows non prize door and u switch prize door u win

p(w2|not w1) = 2/3

p(w2) = p(w2|w1)p(w1)+p(w2|not w1)p(not w1)
= 0*1/3+1*2/3 = 2/3

3> U switch randomly

p(w2) = p(w2/s)p(s)+p(w2/not s)p(not s)
s= switch door randomly
= 2/3*1/2+1/3*1/2 =1/2

clearly switch is good .5>.33 :)

tregor

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hariyo,

why did u have to solve for ith child when u had the answer by grouping...wht would happen if i=3?

can u elaborate plz?

hariyo

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Hi i is the ith choice he makes. Remember he can only make 2 choice(2 choose) only if he doesn't switch he makes only one choice but if he switch he makes other choice i.e. choice 2. As per the question there won't be choice 3. Could you put your solution through grouping ..i would like to see how you solve it.

tregor

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i was talking bt the first problem, hariyo...why did you have to use ith children, i mean can you elaborate a lil on that???

Let's say the car is behind the door Ci and the host reveals the door Rj.

For the player, probabilities p(c1), p(c2) and p(c3) are 1/3 each.

Then, considering the car is in door i (Ci), the host does not have the choice to open the door i (Ri) anymore.

For the host, therefore, p (R1\C1)=0

p (R2\C1)=1/2

p(R3\C1) =1/2

and so on for c2 and c3

Now, If the player initially has chosen door 1, and the third door is revealed (it doesn't matter what assumption you make),

Probability that the car is in door 1 using conditional probability,

p(C1\R3) = P(R3\C1)*P(C1)/P(R3)

=(1/2)* (1/3 )/ ( P(C1R3)+P(C2R3)+P(C3R3) ) {the denominator is equal to P(3)}

=(1/6)/(1/6+1/3+0)

=1/3

Since the third door was revealed, P(c3)=0

that makes P(C2\R3) =1-1/3 =2/3

Sure enough 2/3 is better than 1/3, so you should always switch.

Last edited: 07-Oct-10 12:53 AM

hariyo

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I was making ith child to point it 1st and 2nd child. I think you will get the same result if you try probability for p(w1) too.

tregor

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ya I got it but wondering why you had to calculate P(W2) when the question is not specifically asking about the second. Or I might be missing something , if u could elaborate on those lines..

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