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Math Help - Probability problems

  1. #1
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    Probability problems



    How do i solve these?

    1) Two dice are thrown simultaneously. The probability that the
    product of the two numbers on the two dice is an even number?

    2) A bag contains red and blue marbles totaling between 50 to 100 in number. If two marbles are drawn at random, the chances of them both being red are 25%. One third of the marbles are picked from the bag and thrown away. If one marble is now drawn at random from the bag, what is the probability of it being blue? (Answer in %)

    3) What is the probability that the product of two consecutive non-negative integers will be a number ending with 0?

    ---- My workings:

    For the 1st problem, this is what i did:

    The possible even values on rolling a pair of dice will be 18. So, im getting an incorrect answer of 18/36 (1/2). The source reports 3/4

    For the 2nd one:

    I assumed the set of marbles to be 6, since the answer is supposed to be in % and the 50 to 100 range seems absurd. So, 25% of 6 will be 2 red marbles. Rest are 4 blue marbles. 1/3rd of the total set is removed randomly which makes it obvious that two blue marbles were the ones removed. So, my answer sums up to 50% of blue balls. The answer is correct but am not sure of my logic.

    For the 3rd one:

    I considered the range from 0-9. So, the consecutive numbers ending with 0 are 3 out of 10. So, i am getting 3/10 which is incorrect.
    Last edited by saberteeth; March 2nd 2010 at 06:40 AM.
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  2. #2
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    What have you tried? You'll no doubt get a nice neat answer rolled out from someone, but at least show SOME effort beyond typing out "How do I solve. . ."
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  3. #3
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    Quote Originally Posted by ANDS! View Post
    What have you tried? You'll no doubt get a nice neat answer rolled out from someone, but at least show SOME effort beyond typing out "How do I solve. . ."
    Sorry, i admit that was rude ive now updated this thread with my workings..
    Last edited by saberteeth; March 2nd 2010 at 06:39 AM.
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  4. #4
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    Quote Originally Posted by saberteeth View Post


    How do i solve these?

    1) Two dice are thrown simultaneously. The probability that the
    product of the two numbers on the two dice is an even number?


    ---- My workings:

    For the 1st problem, this is what i did:

    The possible even values on rolling a pair of dice will be 18. So, im getting an incorrect answer of 18/36 (1/2). The source reports 3/4
    18 is too small a count, saberteeth...

    If the number on the first die is even, the product will be even regardless
    of the second die's value.

    You get 18 for that (3 even values for die number 1)(all 6 values for die number 2)

    An odd value on the first die coupled with an even value on the second will also produce an even product.

    How many of those may we have ?

    The final probability can then be calculated

    I guess what you tried to calculate initially was the number of even 2-digit values
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    18 is too small a count, saberteeth...

    If the number on the first die is even, the product will be even regardless
    of the second die's value.

    You get 18 for that (3 even values for die number 1)(all 6 values for die number 2)

    An odd value on the first die coupled with an even value on the second will also produce an even product.

    How many of those may we have ?

    The final probability can then be calculated

    I guess what you tried to calculate initially was the number of even 2-digit values
    Then 3*3*3 = 27. So, 27/36 (3/4) should be the answer. Am i correct?
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  6. #6
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    Yes,

    E(E)=3(3) products Even

    E(O)=3(3) products Even

    O(E)=3(3) products Even

    total is

    3(3)3=3^3
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  8. #8
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    Hello, saberteeth!

    1) Two dice are thrown simultaneously. Find the probability that
    the product of the two numbers on the two dice is an even number.
    There are 6^2=36 possible outcomes.


    The product is odd if both numbers are odd.

    There are 3 choices of an odd number for the first die: \{1,3,5\}
    . . and 3 choices of odd number for the second die.
    Hence, there are: . 3\cdot3\:=\:9 odd-product outcomes.

    So there are: . 36 - 9 \:=\:27 even-product outcomes.


    Therefore: . P(\text{even product}) \:=\:\frac{27}{36} \;=\;\frac{3}{4}




    3) What is the probability that the product of two consecutive
    non-negative integers will be a number ending with 0?

    Two consecutive integers can have the following pairs of units-digits:
    . . (0,1), (1,2), (2,3),(3,4), (4,5), (5,6), (6,7), (7,8), (8,9), (9,0)

    The unit-digits of their product are: . 0,\;2,\;6,\;2,\;0,\;0,\;2,\;6,\;2,\;0


    Got it?

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  9. #9
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    Quote Originally Posted by Soroban View Post
    Hello, saberteeth!

    There are 6^2=36 possible outcomes.


    The product is odd if both numbers are odd.

    There are 3 choices of an odd number for the first die: \{1,3,5\}
    . . and 3 choices of odd number for the second die.
    Hence, there are: . 3\cdot3\:=\:9 odd-product outcomes.

    So there are: . 36 - 9 \:=\:27 even-product outcomes.


    Therefore: . P(\text{even product}) \:=\:\frac{27}{36} \;=\;\frac{3}{4}





    Two consecutive integers can have the following pairs of units-digits:
    . . (0,1), (1,2), (2,3),(3,4), (4,5), (5,6), (6,7), (7,8), (8,9), (9,0)

    The unit-digits of their product are: . 0,\;2,\;6,\;2,\;0,\;0,\;2,\;6,\;2,\;0


    Got it?

    Oh thanks! I had missed (9,0)..
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  10. #10
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    Quote Originally Posted by saberteeth View Post


    How do i solve these?


    2) A bag contains red and blue marbles totaling between 50 to 100 in number. If two marbles are drawn at random, the chances of them both being red are 25%. One third of the marbles are picked from the bag and thrown away. If one marble is now drawn at random from the bag, what is the probability of it being blue? (Answer in %)


    ---- My workings:


    For the 2nd one:

    I assumed the set of marbles to be 6, since the answer is supposed to be in % and the 50 to 100 range seems absurd. So, 25% of 6 will be 2 red marbles. Rest are 4 blue marbles. 1/3rd of the total set is removed randomly which makes it obvious that two blue marbles were the ones removed. So, my answer sums up to 50% of blue balls. The answer is correct but am not sure of my logic.
    Hi saberteeth,

    25% of 8 is 2.

    If half of the marbles were red,
    then the probability of getting 2 reds is \left(\frac{R}{T}\right)\left(\frac{R-1}{T-1}\right)

    where R=number of reds, T=Total number of marbles

    That is very close to \frac{1}{4}
    but not quite 0.25.

    It's approximate.

    That being the case, half the marbles are blue also.

    If one-third of all of the marbles are discarded,
    aproximately the same number of reds and blues will have been discarded.
    This still leaves approximately the same number of reds and blues remaining, however many there may be.

    Hence, there is still a half-chance of choosing a blue on the first go.
    Last edited by Archie Meade; March 3rd 2010 at 01:18 AM.
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  11. #11
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    Hello again, saberteeth!

    Is that the exact wording of problem #2?
    The numbers don't work out . . .


    2) A bag contains red and blue marbles totaling between 50 to 100 in number.
    If two marbles are drawn at random, the chances of them both being red are 25%.

    I can't get past this point . . .

    One third of the marbles are picked from the bag and thrown away.
    If one marble is now drawn at random from the bag,
    what is the probability of it being blue? (Answer in %)

    Let: . \begin{array}{cccc}T &=& \text{total number of marbles:} & 50 < T < 100 \\<br />
R &=& \text{number of Red marbles} \end{array}


    There are: . \begin{Bmatrix}\dfrac{R(R-1)}{2}\text{ ways to draw 2 Red marbles} \\ \\[-3mm]<br />
\dfrac{T(T-1)}{2}\text{ ways to draw }any\text{ 2 marbles} \end{Bmatrix}


    We have: . \frac{\dfrac{R(R-1)}{2}}{\dfrac{T(T-1)}{2}} \:=\:\frac{1}{4}

    . . which simplifies to: . 4R^2 - 4R - T(T-1) \:=\:0


    Quadratic Formula: . R \;=\;\frac{-(\text{-}4) \pm\sqrt{(\text{-}4)^2 - 4(4)[\text{-}T(T-1)]}}{2(4)}

    . . \text{which simplifies to: }\;R \;=\;\frac{1 \pm\sqrt{T^2-T+1}}{2}


    Since R is an integer, the discriminant must be a square.
    But this happens only for: T = 0, 1
    . . which does not allow two red marbles to be drawn.

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