1) 6 boys and 2 girls are seated in a row for a photograph. Then the probability that the girls are not together is:

       A) 1/3

       B) 1/2

       C) 1/4

       D) 3/4

2) The probability that a leap year has 53 sundays is :

       A) 1/7

       B) 2/7

       C) 3/7

       D) 4/7

3) If boys and girls are equally likely, then out of 600 families of 3 children, the number of families having all girls is :

     A) 400

     B) 300

     C) 225

     D) 75

4) One bag contains 4 white and 5 black balls. Another bad contains 6 white and 7 black balls. A ball is transferred from the first bad to the second bag. The probabiliry that the ball drawn is white is :

     A) 4/63

     B) 15/63

     C) 29/63

     D) None of these.

3/4?

1st ka 3/4... second ka 1/7.. working on third and fourth..

For first, consider both girls as a single unit and subtract from total possibilities..

For second, 366 days.. So two days will occur 53 times, rest 52 times. Probability for each day is equal... So 1/7

1) 3/4 is right.



2) answers 2/7.

ofc 3/4 its bcos its (8!-2*7!)/8!

2nd is obviously 2/7 bcos there are 2 days of a new week in a leap year as to one day in a normal year

C:\Documents and Settings\Administrator\My Documents\My Pictures

arrey yaar 2/7 is perfectly correct.

out of 366 days 364 would have 52 sundays , now the 365th and the 366th day can be either of

sm,mt,tw,wt,tf,fs,ss

fav outcomes are 2 and sample space  is 7 so prob = 2/7

yeah.. sorry.. my bad.. :P :P

q1) is fairly simple too..

first place all the 6 boys, now there are 7 gaps b/w the boys the 2 girls are to be placed in these 7 gaps, but the girls can also arrange themselves in 2! ways and the boys in 6! ways so the fav no. of cases are

7C2*2!*6!

prob =

7C2*2!*6! / 8! = 3/4

2) I think it's (b)   2/7