Probability~~ (2 Viewers)

[YannY]

In a game, two players take turns at drawing, and immediately replacing, a marble from a bag
containing two green and three red marbles. The game is won by player A drawing a green marble, or
player B drawing a red marble. Player A draws first. Find the probability that:

i. A wins on her first draw;
ii. B wins on her first draw;
iii. A wins in less than four of her turns;
iv. A wins eventually.¤

The first three questions are easy but the last question iv brings up some questions for me.

My question is what does eventually mean?

The answer is 10/19

From this i concluded that eventually means the chance it can win after the first loss i.e he losses first but what is the possibility that he will win if B will always lose.

Am i right or was what i just said far too ambigious?

 

[Iruka]

How did you answer the third part?

I calculated the probability of A winning within 4 turns by adding the probability that she wins on the first turn + probability she wins the second (obviously, this means that nobody wins the first round) + probability she wins the third turn.

Now if you do this, you will see a pattern emerging (actually, you will see a geometric sequence is emerging). To calculate the probability of A winning eventually (i.e., after an unspecified number of turns), you have to find the limiting sum of this geometric sequence.

 

[Trebla]

This means you add up ALL possibilities:

P(A wins on each draw) = 2/5
P(B does not win on each draw) = 3/5
P(B does not win on each draw) = 2/5

So A can win on 1st, 3rd, 5th draw etc:
P(A wins on 1st draw) = 2/5
P(A wins on 3rd draw) = (3/5)(2/5)(2/5) = (2/5)²(3/5)
P(A wins on 5th draw) = (3/5)(2/5)(3/5)(2/5)(2/5) = (2/5)³(3/5)²
P(A wins on 7th draw) = (3/5)(2/5)(3/5)(2/5)(3/5)(2/5)(2/5) = (2/5)⁴(3/5)³
etc...
Sum them up to add all possible outcomes (call this S):
S = 2/5 + (2/5)²(3/5) + (2/5)³(3/5)² + (2/5)⁴(3/5)³ + .........
This is an infinite sum of a geometric series, with a = (2/5) and r = (3/5)(2/5) so
S = (2/5) / [1 - (2/5)(3/5)]
= (2/5) / [19/25]
= 10/19

 

[YannY]

Ah thanks i got my numbers stuffed up



The third is done by 2/5+(2/5)^2.(3/5)+(2/5)^3.(3/5)^2

Because if he is to win in less than 4 turns he has to win the first three wins and also B has to lose in the first three turns so add up the probability he wins on the first turn, second turn and finally the third turn.

 

[undalay]

i did this question b4. Had 2 ask charles for answersl oool

 

[aakash]

Another way of doing this:
Let probability that A eventually wins = y

y = 2/5 + ( 3/5 * 2/5 * y)
hence, y=10/19

explanation: probability that A eventually wins = P (he wins in 1st trail) + P( A looses in 1st trial) * P(B looses in 1st trial) * P(A eventually wins).

Basically, after both A and B looses in the 1st trial, we are at the "beginning of the process" from where the probability that A eventually wins is simply = y.

 

[aakash]

Another good question:

4000 tickets are sold in a raffle in which there are ten prizes. If you buy 40 tickets, find the probability that:

1) You win 1 prize
2) You win 10 prizes
3) You win no prizes

 

[undalay]

Another way of doing this:
Let probability that A eventually wins = y

y = 2/5 + ( 3/5 * 2/5 * y)
hence, y=10/19

explanation: probability that A eventually wins = P (he wins in 1st trail) + P( A looses in 1st trial) * P(B looses in 1st trial) * P(A eventually wins).

Basically, after both A and B looses in the 1st trial, we are at the "beginning of the process" from where the probability that A eventually wins is simply = y.

that's crazy ) :

 

[aakash]

that's crazy ) :

lol its called recursive method

 

[undalay]

Another good question:

4000 tickets are sold in a raffle in which there are ten prizes. If you buy 40 tickets, find the probability that:

1) You win 1 prize
2) You win 10 prizes
3) You win no prizes

2)
40!3990!
---------------------
4000!30!

3)
(3960)^ 10 x 3990!
------------------------
4000!

how do you do number 1) (in a short way)

 

[qqmore]

Another good question:

4000 tickets are sold in a raffle in which there are ten prizes. If you buy 40 tickets, find the probability that:

1) You win 1 prize
2) You win 10 prizes
3) You win no prizes

Could you use binomial probability?

p(win) = 1/100
p(not win) = 99/100

1) P(win = 1) = 10C1 * (1/100) * (99/100)^9

2) P(win = 10 ) = (1/100)^10

3) P (win = 0 ) = (99/100)^10

 

[undalay]

Could you use binomial probability?

p(win) = 1/100
p(not win) = 99/100

1) P(win = 1) = 10C1 * (1/100) * (99/100)^9

2) P(win = 10 ) = (1/100)^10

3) P (win = 0 ) = (99/100)^10

No because ur assuming replacement

 

[aakash]

I dont think Part 1 can be done in a shorter way...
and i got something else for 3