IBM Probability Quiz -1

first women sit in any of four seats = 4 ways

2nd women sit in any of the 3seats = 3 ways

So total = 4*3 = 12 .

Remaining 6 chairs are there then 1 men can sit in any of 6

2nd can sit in any of 5

3rd can sit in any of 4

6*5*4=120

Then total = 120*12 = 1440

5+4+3+2+1
5 possibilities of couple with the first child
4 for the second
3 for the next
as you remove the couple already used

The average length of three tapes=6800
a+b+c/3=6800
a+b+c=20400
no tape is less than 6400 and we have to find the largest
possibility of one tape,
so, 6400+6400+X = 20400
hence X = 7600

let the no be x
x/50=>r=16
x/80=>r=?
Formula:(Divisor*quotient)+remainder=Dividend.
Soln;
now ,
((50*q)+16)=x----->(1)=>now it is the Dividend for the second condition.
so , x/80=r--------->(2)
sub, q=1 r=66
q=2 r=36
q=3 r=6
q=4 r=56
q=5 r=26
q=6 r=76
q=7 r=46
q=8 r=16
q=9 r=66(starts repeated again)
so,n=8

36C2....
=36*35/2*1

The following are possible combinations of two children
that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls.
Hence the probability is 1/3

within 1 to 100 there are 19 numbers containing digit 2.2
first is 20 to 29 there are 10 numbers and 2 12,32,-,- ------92.
and 200 to 299 there are 100 numbers. so 1100 contain
290 numbers containing digit 2.(100---19 so 100 contain 190
,excluding 200 to 299) 190+100=290.
so 290/1100 is answer.

As the objects are replaced, the probability of drawing red
or blue is equal.
Probability to draw exactly 3 red hats and 1 blue hat
=1/2*1/2*1/2*1/2=1/16
Similarly probability to draw exactly 3 blue hats and
1 red hat =1/2*1/2*1/2*1/2=1/16
Total probability =1/16+1/16=1/8

The first step of the game would be like 8*79. the following step
will fall as 5*46 and the 2nd last step will be like 2*13.
the final value would be 0*(any value). and thus the sum
would be equal to 19.

Their sum can be 3, 4 , 6, 8 ,9 ,12

The probability of getting 3 as sum of two die is = 2/36
The probability of getting 4 as sum of two die is = 3/36
The probability of getting 6 as sum of two die is = 5/36
The probability of getting 8 as sum of two die is = 5/36
The probability of getting 9 as sum of two die is = 4/36
The probability of getting 12 as sum of two die is = 1/36

So, the probability that their sum is a multiple either
of 3 or 4 will be = (2+3+5+5+4+1)/36
= 20/36
=5/9