39
Dr.House
·2009-10-12 09:03:28
1) i am not sure but no other idea strking me as of now.. X be no :of defective articles picked. so we need to find P[X=3]+P[X=4]+P[X=5]+P[X=6] = 3/10C6 +4/10C6+5/10C6+6/10C6 =18/10C6
4
UTTARA
·2009-10-12 09:18:05
Sorry .....
ANSWER for 1st one is 19/42
I think the mistake may be in considering 5 or 6 defective articles since only max 4 r possible
4
UTTARA
·2009-10-12 10:12:13
Hey I got the solution now
P = 6C3 4C3 + 6C2 4C3
6C6 4C0 + 6C5 4C1 + 6C4 4C2 +6C34C3 + 6C2 4C4
= 19 / 42 ANSWER
4
UTTARA
·2009-10-14 07:07:15
Second ans copied from a link given by b555
A random book will be published with probability 1/2 x 2/3 + 1/2 x 1/4 = 11/24
. The probability of him writing two books none of which gets published must therefore be
[1 - (11/24)]2
The probability that he we will get at least one published is therefore:
1 - (1 - 11/24) 2 = 407/576 ANSWER
4
UTTARA
·2009-10-15 00:44:48
More on PROBABILITY
3 ) Two cards are drawn from a well shuffled pack of 52 cards. Find the probability that one of them is a red card & the other is a queen.
4)Each of the 'n' passengers sitting in a bus may get down from it at next stop with probability
P.Moreover, at the next stop either no passenger or exactly one passenger boards the bus. The
probability of no passenger boarding the bus at the next stop being Po . Find the probability that when
the bus continues on its way after the stop, there will again be 'n' passengers in the bus.