[doomsayer]

Originally Posted by CrackCatWalkIn

The latest question on Cat Fundae

--->
What is the probability that a man having four children has two girls and two boys, given that he has at least one girl.

The answer choices are a) 0.375 b) 0.5 c) 0.4 d) None of these
--->

ans is (C) .4

6/16 /15/16 = 2/5 = .4

[aravindva]

Originally Posted by CrackCatWalkIn

The latest question on Cat Fundae

--->
What is the probability that a man having four children has two girls and two boys, given that he has at least one girl.

The answer choices are a) 0.375 b) 0.5 c) 0.4 d) None of these
--->

Answer is 0.25

[MITS@BITS]

: Concepts...total fundas!! - 06-09-2007, 11:33 AM
Quote:
Originally Posted by MITS@BITS (Concepts...total fundas!!)
ans is --->1/5 (b)

nswer is right.... but how come??

just create quadratic equation for 4 (for a, b, c ,d)--in their respective terms
nd start frm d's equation--1 more condition--for real ans, check discriminent, it is 0, roots should be equal.
25.d^2-40.d+16=0-------(1)gives d=4/5
50.c^2-60.c+18=0-------(2)gives c=3/5
similarly make equation for b & a
got values b=2/5
a=1/5
a,b,c,d are in AP with common difference 1/5

We can also solve this question by (a+b+c+d)^2

[alohabrij]

Originally Posted by aravindva

Answer is 0.25

hi aravindva,

theres one more method which seems correct but the ANS differs , please check

The probablity of second child being a girl and other two being boys is 1/2*1/2*1/2
similarly the probab of third being a girl and rest two guys is 1/2*1/2*1/2
and last being a girl is again 1/2*1/2*1/2

so the ans turns out to be .375
:

[macora]

Originally Posted by CrackCatWalkIn

The latest question on Cat Fundae

--->
What is the probability that a man having four children has two girls and two boys, given that he has at least one girl.

The answer choices are a) 0.375 b) 0.5 c) 0.4 d) None of these
--->

The options can be
1g 3b
2g 2b
3g 1b
4g 0b

so, 0.25 and so none of these

[esh.nil]

Originally Posted by alohabrij

hi aravindva,

theres one more method which seems correct but the ANS differs , please check

The probablity of second child being a girl and other two being boys is 1/2*1/2*1/2
similarly the probab of third being a girl and rest two guys is 1/2*1/2*1/2
and last being a girl is again 1/2*1/2*1/2

so the ans turns out to be .375
:

@BRIJESH.... I feel that the answer is 1/4....
and the events are independent of each other(mutually exclusive)... the occurance of a girl or boy child at anypoint is independent of previous occurances.... so u cant multiply the probabilities...

[garggaura]

Originally Posted by macora

The options can be
1g 3b
2g 2b
3g 1b
4g 0b

so, 0.25 and so none of these

yes it shd be 0.25.
BBBB
GGGG
BBBG
BBGG
GGGB

out of above 5 possiblities first one will be ruled out (atleast 1 girl)
so 1/4 = 0.25

n'joy
Gaurav

[laxmii]

hey,this is laxmi
I plan to attempt 4 CAT2008
CAN u pls suggest me a good coaching????
reply

[sachin2807]

Q> For any natural number q,the sum [(3q+1)/2] + [(3q+2)/4] +........+[(3q+2^p)/2^(p+1)]; where p is such that 3q<2^p and [x] = greatest integer less than or equal to x,is equal to:

a) q b) 3q c) 2q d) None of these

[MITS@BITS]

Q> For any natural number q,the sum [(3q+1)/2] + [(3q+2)/4] +........+[(3q+2^p)/2^(p+1)]; where p is such that 3q<2^p and [x] = greatest integer less than or equal to x,is equal to:

a) q b) 3q c) 2q d) None of these

Ans. is (b)---3q
we can write this in the form of two series(both are GP)
1st GP [3q/2]+[3q/4]+---------+[3q/2^(p+1)]
2nd GP [1/2]+[2/4]+---------+[2^p/2^(p+1)]
sum of first GP with common ratio 1/2 is [3q.((2^p)+1/2^p)]-------(1)

sum of second GP is 0-------(2)
frm equation (1)
[3q+(3q/2^p)]
it is given in question that
3q<2^p i.e. 3q/2^p<1--------(3)
frm (3)
ans is 3q