Thread: Probability question

Hey guys I cant do this following question so thanks in advance if you could help!

A bag contains 2 red marbles, 5 white marbles and 3 black marbles. A game is played with a player randomly drawing one marble at a time from the bag with replacement, continuing until he obtains a red marble or a white marble. He wins if the last marble is white and loses otherwise.

(a) Find the probability that the player wins.
(b) Find the minimum value of n for which P(he draws at most n marble to win) is more than or equal to 0.7.

Hello, margaritas!

I can help with part (a) . . . I'm still working on part (b).

A bag contains 2 red marbles, 5 white marbles and 3 black marbles. .A game is played
with a player randomly drawing one marble at a time from the bag with replacement,
continuing until he obtains a red marble or a white marble.
He wins if the last marble is white and loses otherwise.

(a) Find the probability that the player wins.

We have the following probabilitues:
. . P(W) = 0.5 . wins
. . P(R) = 0.2 . loses
. . P(B) = 0.3 . draws again


Let's baby-step through this . . .

He could win on the first draw: .P(W1) .= .0.5

He could win on the second draw.
. . He must draw B on the first draw: P(B1) = 0.3
. . then draw W on the second draw: P(W2) = 0.5
Hence: .P(W2).= .(0.5)(0.3)

He could win on the third draw.
. . He must draw B on the first draw: P(B1) = 0.3
. . then draw B on the second draw: P(B2) = 0.3
. . then draw W on the third draw: P(W3) = 0.5
Hence: .P(W3) .= .(0.5)(0.3)²

and so on . . .


Then the probability that he wins is:
. . P(win) .= .0.5 + (0.5)(0.3) + (0.5)(0.3)² + (0.5)(0.3)³ + . . .

This is a geometric series with first term a = 0.5 and common ratio r = 0.3

. . . . . . . . . . . . . . .0.5 . . . . 0.5
Its sum is: .S .= . --------- .= .----
. . . . - . . . . . . . . 1 - 0.3 . - . 0.7

Therefore: .P(win) .= .5/7

Originally Posted by margaritas

Hey guys I cant do this following question so thanks in advance if you could help!

A bag contains 2 red marbles, 5 white marbles and 3 black marbles. A game is played with a player randomly drawing one marble at a time from the bag with replacement, continuing until he obtains a red marble or a white marble. He wins if the last marble is white and loses otherwise.

(a) Find the probability that the player wins.
(b) Find the minimum value of n for which P(he draws at most n marble to win) is more than or equal to 0.7.

(a) Since the marbles are replaced after each draw the probability on a given
draw of a play ending in a win is constant, as is the probability of a loss. So of those
games which end on any draw 5/7 are wins and 2/7 are loses. As the game
will eventualy end (with probability 1) 5/7 is the probability that it ended in
a win.

RonL

Thanks so much Soroban and CaptainBlack, I understand part (a) now but still cant get part (b) though!

Originally Posted by margaritas

Hey guys I cant do this following question so thanks in advance if you could help!

A bag contains 2 red marbles, 5 white marbles and 3 black marbles. A game is played with a player randomly drawing one marble at a time from the bag with replacement, continuing until he obtains a red marble or a white marble. He wins if the last marble is white and loses otherwise.

(a) Find the probability that the player wins.
(b) Find the minimum value of n for which P(he draws at most n marble to win) is more than or equal to 0.7.

(b) the probability that the game ends with a win on the r-th draw is:

p(r) = (3/10)^(r-1) (5/10)

That is the game must not finish on the first r-1 draws, and must end
with a win on the r-th draw.

P(he draws at most n marble to win) = sum_{r=1 to n} p(r)

.......= (5/10) [1+(3/10) + (3/10)^2 + ... + (3/10)^{n-1}]

.......= (5/10)[1+(3/10)^{n}]/[1-(3/10)]

Now by trial and error we find P(he draws at most 3 marble to win)~=0.695
and P(he draws at most 4 marble to win)~= 0.7085, so the minimum n such
that P(he draws at most n marble to win)>=0.7 is 4.

(By the way all of this is implicit in Soroban's solution - the main problem that
I had with it was working out what the h*** it was asking)

RonL

Hello again, Margaritas!

(b) Find the minimum value of n for which:
. . .P(he draws at most n marble to win) > 0.7

The probability that he wins in n draws is:

P(n draws) .= .0.5 + (0.5)(0.3) + (0.5)(0.3)² + (0.5)(0.3)² + . . . + (0.5)(0.3)^{n-1}


This is a geometric series with first term a = 0.5, common ratio r = 0.3, and n terms.

. . . . . . . . . . . . 1 - 0.3^n . . . .5(1 - 0.3^n)
Its sum is: .(0.5) ----------- . = . --------------
. . . . . . . . . . . . . 1 - 0.3 . . . . . . . . .7


. . . . . . . . 5(1 - 0.3^n)
We have: .---------------- . > . 0.7
. . . . . . . . . . . 7


Multiply by 7/5: . 1 - 0.3^n . > . 0.98

Subtract 1: . -0.3^n . > . -0.02

Divide by -1: . 0.3^n . < . 0.02 .*

Take logs: . ln(0.3^n) . < . ln(0.02)

We have: . n·ln(0.3) . < . ln(0.02)


. . . . . . . . . . . . . . . . . . .ln(0.02)
Divide by ln(0.3): . n . > . ---------- . = . 3.249261936 .**
. . . . . . . . . . . . . . . . . . . ln(0.3)

Therefore: .n .> .4

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*
When multiplying or dividing by a negative quantity, reverse the inequality.

**
Note that ln(0.3) is negative.

Thanks I get it now!