[rajbahadur tomar]

A bug crawls along the edges of a regular tetrahedron ABCD with edges length 1. It starts at A and at each vertex chooses its next edge at random (so it has a 1/3 chance of going back along the edge it came on, and a 1/3 chance of going along each of the other two). Find the probability that after it has crawled a distance 7 it is again at A is p.

[maddevil]

Originally Posted by rajbahadur tomar

A bug crawls along the edges of a regular tetrahedron ABCD with edges length 1. It starts at A and at each vertex chooses its next edge at random (so it has a 1/3 chance of going back along the edge it came on, and a 1/3 chance of going along each of the other two). Find the probability that after it has crawled a distance 7 it is again at A is p.

i guess there are ample threads already running for quant, plz post ur question on one of them.

D.M.

P.S: No offence meant

[rajbahadur tomar]

A deck of 40 cards has four each of cards marked 1, 2, 3, ... 10. Two cards with the same numbe are removed from the deck. Find the probability that two cards randomly selected from the remaining 38 have the same number as each other.

[khanna_sumit]

Originally Posted by rajbahadur tomar

A deck of 40 cards has four each of cards marked 1, 2, 3, ... 10. Two cards with the same numbe are removed from the deck. Find the probability that two cards randomly selected from the remaining 38 have the same number as each other.

if i have interpreted it correctly then that means that we know that two particular cards have been removed lets say two 1s...in that case there are 2 1s and 4 cards of 2s, 3s, .....10s are there in the deck.....so the answeris {C(4,2)*9+C(2,2)}/C(38,2)......

[ankursanghi]

Originally Posted by khanna_sumit

if i have interpreted it correctly then that means that we know that two particular cards have been removed lets say two 1s...in that case there are 2 1s and 4 cards of 2s, 3s, .....10s are there in the deck.....so the answeris {C(4,2)*9+C(2,2)}/C(38,2)......

khanna sahib,

could you please explain, for a probability-phobic guy, how u achieved the answer??

rgds,
ankur

[khanna_sumit]

Originally Posted by ankursanghi

khanna sahib,

could you please explain, for a probability-phobic guy, how u achieved the answer??

rgds,
ankur

after removing two cards which are same( lets say two 1s)
we are left with 36 cards ( 2,3,4,5,6,7,8,9,10 ) i.e. 4 cards of each type and two cards numbered 1......now if a draw 2 cards and want those two cards to be of same type then either both are 1s or both 2s and so on.......out of two 1s, we can select two 1s in C(2,2) ways....out of four 2s we can select two 2s in C(4,2) ways...similarly number of ways of selecting cards numbered (3-10) are C(4,2) each.....hence total number of ways of selecting 2 same cards from these 38 cards are C(4,2)+C(4,2).....9 times + C(2,2) = 9*C(4,2)+C(2,2)
..total number of ways of selcting 2 cards from a set of 38 cards are C(38,2)...

so prob = favourable cases/total cases =
{9*C(4,2)+C(2,2)}/C(38,2)

rgds

[khanna_sumit]

Originally Posted by rajbahadur tomar

A bug crawls along the edges of a regular tetrahedron ABCD with edges length 1. It starts at A and at each vertex chooses its next edge at random (so it has a 1/3 chance of going back along the edge it came on, and a 1/3 chance of going along each of the other two). Find the probability that after it has crawled a distance 7 it is again at A is p.

is the answer 7/48? just solved it mentally. not sure....but give some time....wanna solve it myself first...

rgds

[khanna_sumit]

Originally Posted by rajbahadur tomar

A bug crawls along the edges of a regular tetrahedron ABCD with edges length 1. It starts at A and at each vertex chooses its next edge at random (so it has a 1/3 chance of going back along the edge it came on, and a 1/3 chance of going along each of the other two). Find the probability that after it has crawled a distance 7 it is again at A is p.

I got the answer....it is 182/729.

[made_for_iims]

Originally Posted by khanna_sumit

I got the answer....it is 182/729.

seems correct!!!!.....

check the explanation

Key Pt.....The prob of A at particular distance.....will become sure-shot(prob 1) for (B,C,D cluster) in next distance.....:-)

Distance Prob of A prob of rest (B,C,D)

0 1 0
1 0 1
2 1/3 2/3
3 2/3 * 1/3= 2/9 7/9
4 7/9 * 1/3 = 7/27 20/27
5 20/27 * 1/3 = 20/81 61/81
6 61/243 182/243
7 182/243 * 1/3 = 182/729 547/729


Prob of BCD can be calculated from (1-A) & also frm the fact

prob of A(n-1) + 2/3 prob of B (n - 1 )

:-)

[khanna_sumit]

Originally Posted by made_for_iims

seems correct!!!!.....

check the explanation

Key Pt.....The prob of A at particular distance.....will become sure-shot(prob 1) for (B,C,D cluster) in next distance.....:-)

Distance Prob of A prob of rest (B,C,D)

0 1 0
1 0 1
2 1/3 2/3
3 2/3 * 1/3= 2/9 7/9
4 7/9 * 1/3 = 7/27 20/27
5 20/27 * 1/3 = 20/81 61/81
6 61/243 182/243
7 182/243 * 1/3 = 182/729 547/729


Prob of BCD can be calculated from (1-A) & also frm the fact

prob of A(n-1) + 2/3 prob of B (n - 1 )

:-)

i solved it using a different method in which i calculated the number of ways in which this can happen.......used a big sheet of paper for it though....but it was really interesting problem......ur solution is quite systematic.......good approach.

rgds