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    hi im doing an application and it asks me this question...

    Three fair dice are thrown. What is the probability the exactly two of the scores
    are sixes?

    On my way to school I pass through two sets of traffic lights that operate
    independently. The probabilities that I have to wait at these two sets of traffic
    lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

    A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?



    im not good at this type of maths so please help me clearly.

    thank you!
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    (Original post by Student#123)
    ...
    What have you tried?

    Did you make the questions up or is that how they are written?
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    Do you mean roll three dice and you get a double six?

    P(6) P(6') does this mean anything to you?
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    (Original post by SubAtomic)
    Do you mean roll three dice and you get a double six?

    P(6) P(6') does this mean anything to you?
    Three fair dice are thrown. What is the probability the exactly two of the scores
    are sixes?


    For this question I did

     P(exactly two sixes)=P(6) \times P(6)   \timesP(6')=(\frac{1}{6})^2 \times \frac{5}{6}=\frac{5}{216} which is incorrect.


    On my way to school I pass through two sets of traffic lights that operate
    independently. The probabilities that I have to wait at these two sets of traffic
    lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

    P(delay on 1st and 2nd)=0.3 \times 0.4=0.12

    p(delayed by at least 1 traffic light)=1-(0.12)^2=0.9856 which is incorrect.


    A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?

    I attempted a tree diagram for this question but it got messy and confused me.
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    1/6*1/6 if Im not mistaken


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    (Original post by nawfaall)
    1/6*1/6 if Im not mistaken


    Posted from TSR Mobile
    that's wrong too.
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    the answer is


    3*5/(6^3)
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    (Original post by Student#123)
    Three fair dice are thrown. What is the probability the exactly two of the scores
    are sixes?


    For this question I did

     P(exactly two sixes)=P(6) \times P(6)-P(6')=(\frac{1}{6})^2 \times \frac{5}{6}=\frac{5}{216} which is incorrect.
    Why did you do - P(6') ? The three dice will show something so it is multiply all the way. P(6) and P(6) and P(6'). 5/216 is correct. Unless it is asking for the sum of two dice to be six :s. The question is badly written.

    (Original post by Student#123)
    On my way to school I pass through two sets of traffic lights that operate
    independently. The probabilities that I have to wait at these two sets of traffic
    lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

    P(delay on 1st and 2nd)=0.3 \times 0.4=0.12

    p(delayed by at least 1 traffic light)=1-(0.12)^2=0.9856[/tex] which is incorrect.
    Add up all the possibilities, both traffic lights could delay you, or one could or the other could.


    (Original post by Student#123)
    A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?

    I attempted a tree diagram for this question but it got messy and confused me.
    P(HHH)= (0.47)^3

    Does that help?
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    (Original post by SubAtomic)
    Why did you do - P(6') ? The three dice will show something so it is multiply all the way. P(6) and P(6) and P(6'). 5/216 is correct.



    Add up all the possibilities, both traffic lights could delay you, or one could or the other could.




    P(HHH)= (0.47)^3

    Does that help?
    5/216 is not correct.

    I have to choose from:

    (b) 5/72
    (c) 25/72
    (d) 25/216




    . A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?
    (a) Two tails and one head
    (b) Two heads and one tail
    (c) Three heads
    (d) Three tails
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    (Original post by Student#123)
    hi im doing an application and it asks me this question...

    Three fair dice are thrown. What is the probability the exactly two of the scores
    are sixes?

    On my way to school I pass through two sets of traffic lights that operate
    independently. The probabilities that I have to wait at these two sets of traffic
    lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

    A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?



    im not good at this type of maths so please help me clearly.

    thank you!

    the answer is being 3(5)/6^3

    there are 6^3 total possible out come

    there are three possible outcome of 6, 6, non six

    each non six can be five possibility

    so answer is 15/216




    EDIT: 15/216 = 5/72
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    (Original post by Mullah.S)
    the answer is being 3(5)/6^3

    there are 6^3 total possible out come

    there are three possible outcome of 6, 6, non six

    each non six can be five possibility

    so answer is 15/216
    thank you
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    for second question:



    0.3 and 0.4 probability delay


    total options are equal:


    delay - delay (0.3*0.4)

    delay - non delay (0.3*0.6)

    non delay - delay (0.7*0.4)

    non delay - non delay (0.7*0.6)


    total probability of 1 delay = (0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)
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    (Original post by SubAtomic)
    Why did you do - P(6') ? The three dice will show something so it is multiply all the way. P(6) and P(6) and P(6'). 5/216 is correct. Unless it is asking for the sum of two dice to be six :s. The question is very badly written.



    Add up all the possibilities, both traffic lights could delay you, or one could or the other could.




    P(HHH)= (0.47)^3

    Does that help?
    I get 0.53 for the traffic light question.

    A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?
    (a) Two tails and one head
    (b) Two heads and one tail
    (c) Three heads
    (d) Three tails
    (e) I don’t know

    I thought 3 tails since this has larger probability bit is wrong... could you explain please?
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    (Original post by Mullah.S)
    for second question:



    0.3 and 0.4 probability delay


    total options are equal:


    delay - delay (0.3*0.4)

    delay - non delay (0.3*0.6)

    non delay - delay (0.7*0.4)

    non delay - non delay (0.7*0.6)


    total probability of 1 delay = (0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)

    I have answer as 0.58

    A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?
    (a) Two tails and one head (b) Two heads and one tail
    (c) Three heads (d) Three tails
    (e) I don’t know
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    (Original post by Student#123)
    5/216 is not correct.

    I have to choose from:

    (b) 5/72
    (c) 25/72
    (d) 25/216
    Assuming is fraught with peril. 3 fair dice are thrown and the sum of exactly two dice is six then.
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    3rd question:


    possible outcomes:

    three H (1 time), two H one T (3 times), two T one H (3 times), three T (1 time)

    probabilities

    0.47^3 or 3(0.47^2)(0.53) or 3(0.53^2)(0.47) or 0.53^3


    which one of this 4 number is being biggest number?
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    (Original post by SubAtomic)
    Whoever wrote that question needs a slap. Either 3 fair dice are thrown and the sum is 6 or the sum of two dice is six then.





    You'll need to go through the possible outcomes using the method I showed. However it should be obvious.

     P(HHT) = P(THH) \neq P(TTH)

    you are needing to learn some amateur statistic (like me). do looking at my answer for correction.
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    (Original post by Student#123)
    I have answer as 0.58

    A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
    outcome?
    (a) Two tails and one head (b) Two heads and one tail
    (c) Three heads (d) Three tails
    (e) I don’t know

    you need to put this in calculator

    (0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)
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    (Original post by Mullah.S)
    you are needing to learn some amateur statistic (like me). do looking at my answer for correction.
    i understand this question now. thank you.

    is 0.58 correct for my 2nd question i post?
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    (Original post by Mullah.S)
    you need to put this in calculator

    (0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)
    ok. what formula did you use?
 
 
 
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