four tickets marked 00, 01,10, 11 respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time . the probability that the sum of the numbers on tickets thus drawn is 23?

Question

shubhamsrg · Answer

23 = 10 + 11 + 1 + 1 + 0
     = 10 + 10 + 1 + 1 + 1
     =  11 + 11 + 1 + 0 + 0

are any other combos possible ?

yrelhan4 · Answer

wellthe answer is 25/256. and total number of ways are 1024. so there are 100 favourable ways. :P

shubhamsrg · Answer

I haven't given the ans yet, am just saying 23 can be produced in these 3 ways right,?

shubhamsrg · Answer

the ans would be

(1/4)^5 *5!/2! ( 1 + 1/3! + 1/2!)

shubhamsrg · Answer

=25/256 B|

yrelhan4 · Answer

oh. yeah those are possible combos. 
and how did you get that?

shubhamsrg · Answer

see there are 5 draws, for each draw, probab for any number is same i.e. 1/4

shubhamsrg · Answer

1st combo = 10,11,1,1,0
we can get these in 5! /2! ways

next =10,10,1,1,1  => 5!/2! 3!

last = 11,11,1,0,0 => 5!/2! 2!

net ptobab = (1/4)^5 *  5! /2!      +    (1/4)^5 * 5!/2!3!       + (1/4)^5 * 5!/2! 2!

yrelhan4 · Answer

hmm. got it. thank you.

shubhamsrg · Answer

glad to help.