probability problem!?

Question

Each of (n) urns contains three balls numbersd 1,2 and 3.
a ball is chosen at random from each urn, yielding a group of integers. Find the probability that this group contains each of the integers 1,2 and 3 at least once.

Answer 1

Dear Me Myself And ..........?,

Consider the situation when a ball has just been chosen from urn k. At this point there are three possibilities for the resulting group of integers:
(i) the balls drawn so far are all of one kind,
(ii) the balls drawn so far are of two kinds, or
(iii) the balls drawn so far are of all three kinds.
The probability of the third case, with the group of balls having integers 1, 2, and 3 at least once, is what we are ultimately interested in knowing, but we must consider the other two cases as intermediate stages.

Let f(j,k) be the probability that the group of balls is of j kinds after drawing from urn k.

We can write a set of recursive equations for k > 1 corresponding to our three cases as follows.

f(1,k) = 1/3 f(1,k-1),

f(2,k) = 2/3 (f(1,k-1) + f(2,k-1)),

f(3,k) = 1/3 f(2,k-1) + f(3,k-1),

and the initial condition representing k = 1 having f(1,1) = 1,
f(2,1) = 0, and f(3,1) = 0, since the ball drawn from the first urn must always be of only one kind.

The probability of interest, f(3,n), now can be determined by successive substitutions of our recursive equations for the probabilities at earlier stages n-1, n-2, n-3, . . . , 1. The resulting expression will then be in terms of the initial conditions, immediately giving a numerical result. For example, here is the first substitution:

f(3,n) = 1/3 f(2,n-1) + f(3,n-1)
= 1/3 [2/3 (f(1,n-2) + f(2,n-2))] + [1/3 f(2,n-2) + f(3,n-2)]
= 2/9 f(1,n-2) + 2/9 f(2,n-2) + 1/3 f(2,n-2) + f(3,n-2)
= 2/9 f(1,n-2) + 5/9 f(2,n-2) + f(3,n-2).

If n = 3, then the equation above relates directly to the initial conditions:

f(3,3) = 2/9 f(1,1) + 5/9 f(2,1) + f(3,1)
= 2/9 (1) + 5/9 (0) + (0)
= 2/9.

For larger values of n you could perform similar substitutions, successively obtaining equations for f(3,n) in relation to the initial conditions. By doing that, you obtain the probability that the group of chosen balls contains the integers 1, 2, and 3 at least once.

Answer 2

I NEeD THE TEN bad--SO I THOUGHT this one outONEOUT AND MY ANSWER IS.........27% CHANCE OF GETTIN THE ALL HEARTS IN SEQUENCE FROM AN ACE TO 10. THIS IS HOW I ARRIVED at my decision, ok???? listen well cause i just madeit to level 3 after too long of a time and now i am attempting level 4. if i had 52 cards which are in a deck of cards and u dealt me five like u said u did--that would leave me with 47 cards. minus a ten would equal 37%and the odds on that are even worse-r than that so add another or sub-track another 10 for the odds bein against me and i just ended up with a 27% and am i rite or rong??? i am a kia

Answer 3

so i do take the balls from one urn or from any of the urns. if so i need to know how many there are..

Answer 4

Probability of happening the event = i - probability of not happening the event.

Probability of not happening the event is ( 2/3) ^n.

Therefore the required probability = 1 - (2/3) ^n