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Math Help - Probability question

  1. #1
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    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.
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  2. #2
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    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)▓

    and so on . . .


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

    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

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  3. #3
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    Quote Originally Posted by margaritas View Post
    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
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  4. #4
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    Thanks so much Soroban and CaptainBlack, I understand part (a) now but still cant get part (b) though!
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  5. #5
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    Quote Originally Posted by margaritas View Post
    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
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  6. #6
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    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)▓ + (0.5)(0.3)▓ + . . . + (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: . nĚln(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.

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  7. #7
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    Thanks I get it now!
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