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Math Help - Probability question~

  1. #1
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    Probability question~

    I failed to understand and to visualize the following questions:

    (A little help would be thankful...)


    • A fair die has its 4-spot changed to 5-spot and its 2-spot changed to a 3-spot. Find the probability of getting an even number when the altered die is rolled.

    My assumption here is that it probably won't change the probability before it was altered since it is a fair die and that it has equal chances. (But I was wrong). Then in the end, I might not actually understood the part where it changes spot.

    • A bag contains 5 red cubs and 3 black cubes. Three cubes are chosen at random. Find the probability of at least 2 reds being chosen, given that the first cube was red: i.) if the cubes are replaced after each draw ii.) if the cubes are not replaced after each draw.

    I'm not really sure what they meant by 'given that the first cube was red'
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  2. #2
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    Quote Originally Posted by nabbiechan View Post
    I failed to understand and to visualize the following questions:

    (A little help would be thankful...)


    • A fair die has its 4-spot changed to 5-spot and its 2-spot changed to a 3-spot. Find the probability of getting an even number when the altered die is rolled.

    My assumption here is that it probably won't change the probability before it was altered since it is a fair die and that it has equal chances. (But I was wrong). Then in the end, I might not actually understood the part where it changes spot.

    • A bag contains 5 red cubs and 3 black cubes. Three cubes are chosen at random. Find the probability of at least 2 reds being chosen, given that the first cube was red: i.) if the cubes are replaced after each draw ii.) if the cubes are not replaced after each draw.

    I'm not really sure what they meant by 'given that the first cube was red'
    (1) At first, the spots are 1,2,3,4,5,6. After the change, the spots are 1,3,3,5,5,6.

    P(even number)=...

    (2)(a) Perhaps it will help by drawing a tree diagram. This is an example of a conditional probability.

    Let R be the event of the first cube drawn is red and Q the event of drawing at least 2 red cubes.

    P(Q n R)=P(RRB)+P(RBR)+P(RRR)

    P(R)=P(RRR)+P(RRB)+P(RBR)+P(RBB)

    P(Q|R)=P(R n Q)/P(R)

    (b) You can do the same here but this time its drawing without replacement.
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  3. #3
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    Oh by the way, regarding the second question with b... How will you draw it without the replacement?

    I was actually trying to solve it through combinations as well. but I got frustrated with so much to calculate.
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  4. #4
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    Quote Originally Posted by nabbiechan View Post
    Oh by the way, regarding the second question with b... How will you draw it without the replacement?

    I was actually trying to solve it through combinations as well. but I got frustrated with so much to calculate.
    Yeah, initially i thought of solving this using combinations too until i saw the conditions. The problem here is when you calculate P(getting 2 red cubes)=(5C2 x 3C1)/(8C3) using combinations this accounts for all possible combinations: P(RRB) , P(RBR) , P(BRR)
    where P(BRR) is not required because it doesn't start with a red colour. It's tedious to check all so it's easier to draw a tree diagram.
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  5. #5
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    Question

    Okay. This is what I did based on what you have told me

    (Attempt #5)

    P(Q n R) = P(RRB) + P(RBR) + P(RRR)
    = (5C2 x 3C1)/(8C3) + (5C2 x 3C1)/(8C3) + (3C3)/(8C3)
    = (15/28) + 15/28 + 1/56
    = 61/56

    P(R) = P(RRR) + P(RRB) + P(RBR) + P(RBB)
    = (5C2 x 3C1)/(8C3) + (5C2 x 3C1)/(8C3) + (3C3)/(8C3) + (5C1 x 3C2)/(8C3)
    = 1/56 + 15/28 + 15/28 + 15/56
    = 19/14

    P (Q|R) = P(R n Q) / P(R)
    = (61/56)/(19/14)
    = 61/76

    Correct me if I'm wrong
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  6. #6
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    Quote Originally Posted by nabbiechan View Post
    Okay. This is what I did based on what you have told me

    (Attempt #5)

    P(Q n R) = P(RRB) + P(RBR) + P(RRR)
    = (5C2 x 3C1)/(8C3) + (5C2 x 3C1)/(8C3) + (3C3)/(8C3)
    = (15/28) + 15/28 + 1/56
    = 61/56

    P(R) = P(RRR) + P(RRB) + P(RBR) + P(RBB)
    = (5C2 x 3C1)/(8C3) + (5C2 x 3C1)/(8C3) + (3C3)/(8C3) + (5C1 x 3C2)/(8C3)
    = 1/56 + 15/28 + 15/28 + 15/56
    = 19/14

    P (Q|R) = P(R n Q) / P(R)
    = (61/56)/(19/14)
    = 61/76

    Correct me if I'm wrong
    That's not correct. Remember the condition given here is the first ball drawn MUST be a red. If you use combinations, it doesn't guarantee the first ball drawn to be red.
    Try drawing a tree diagram. Also note that probability of an event must be in the range of 0<p<1.

    P(Q n R) = P(RRB) + P(RBR) + P(RRR)

    =(5/8)(4/7)(3/6)+(5/8)(3/7)(4/6)+(5/8)(4/7)(3/6)

    =15/28

    P(R)=P(RRR)+P(RRB)+P(RBR)+P(RBB)

    =15/28+P(RBB)

    =15/28+(5/8)(3/7)(4/6)

    =5/8

    so P(Q|R)=(15/28)/(5/8)=6/7
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  7. #7
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    Oooooh. Thank you so much! It's all becoming clear to me. I just had this assumption that I have to know how to solve it through combinations but oh well. Thanks again!

    EDIT: I believe you made a small mistake with P(R) on 4/6. It's supposed to be 2/6 otherwise it will become 5/7~
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  8. #8
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    Quote Originally Posted by nabbiechan View Post
    Oooooh. Thank you so much! It's all becoming clear to me. I just had this assumption that I have to know how to solve it through combinations but oh well. Thanks again!

    EDIT: I believe you made a small mistake with P(R) on 4/6. It's supposed to be 2/6 otherwise it will become 5/7~
    yeah, that's a mistake and you are welcome.
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