1wat r the options........mani.??........
Here i will be posting ques on probability which could prove to be useful 4 every1
Do try to give explanations if u can
options will be given on demand[1]
Q1
If the papers of 4 students can be checked by any one of 7 teachers , then the probability that all the 4 papers r checked by exactly 2 teachers
-
UP 0 DOWN 0 0 80
80 Answers
11Q2
2 numbers r selected at random 4m 1,2,3,4,5,........100 and r multiplied
then the probability(correct to 2 place of decimals ) that the product obtained id divisible by 3
11yes it was a mystery
sorry 4 confusing u guys [2]
but sacchi i didnt meant to
1dats what me n tapan said... 42/343 = 6/49...
well, none of these is correct... rather... E option is correct... [3]
11Now i am disclosing the solution
I think no one thought like that
the total cases r 74 everyone knows it
the no. of ways of of choosing 2 teachers out of 7 are 7C2
but the no. of ways they can check 4 papers is 24
But it includes 2 ways in which paper will be selected by 1 teacher
So no. of ways become 24-2=14
No. of fav ways 7C2 X14
the prob is 7C2 X14/74 =6/49
so u see which option is correct!!!!!!!!!!!
11WRONG PROCEDURE
I KNOW TUKKE MEIN AAP MAHIR HO
PLEASE GIVE THE RIGHT PROCEDURE!!!!!!!!!!!
1[11]
can u tell us how u interpret this...!!??? [7]
btw... its value is greater dan one... so its not possible...!!!
1Answer is 42/343... from d expression given by tapan, which according to me is correct...!!!
well, whats d answer given...???
11the options r
(A)2/7
(B)12/49
(C)32/343
(D)NONE OF THESE
NOW I DONT WANT THE ANSWER
NOW I WANT THE SOLUTION
TUKKE MAR KE ANS TO TUM NIKAAL HI LOGE
1@singh...
if all d papers n teachers are different... then second one is almost appropriate... if they are considered identical... then v need to further solve it...!!!