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Thread: coin tossed

  1. #1
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    coin tossed

    a coin is tossed 3 times. If both heads and tails appear find the probability that exactly one tail appears.

    i am using the conditional probability rule so p(probability of getting one tail | probability of getting one tail and one head )=
    p(probability of getting one tail intersection probability of getting one tail and one head)/p(probability of getting one tail and one head)

    i am having trouble finding the probability of getting one tail and one head as it could be in any order so i cant say (.5)^3
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  2. #2
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    Re: coin tossed

    Quote Originally Posted by markosheehan View Post
    a coin is tossed 3 times. If both heads and tails appear find the probability that exactly one tail appears.
    Toss a coin three times. That produces eight elementary outcomes. By the conditional, we eliminate all heads & all tails. Leaving six outcomes. Of those six only three consists of two heads and one tail. Thus what is the probability?
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  3. #3
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    Re: coin tossed

    In case you're too busy:
    head=1, tail=2
    111
    112
    121
    122
    211
    212
    221
    222
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  4. #4
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    Re: coin tossed

    1/2 laying it out on a tree diagram helped me.
    its .375/.75 =.5
    Last edited by markosheehan; Jun 26th 2017 at 01:15 AM.
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  5. #5
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    Re: coin tossed

    Geesh Marko...keep it simple!

    head=1, tail=2
    111 n/a
    112 *
    121 *
    122
    211 *
    212
    221
    222 n/a

    3/6
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  6. #6
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    Re: coin tossed

    Quote Originally Posted by DenisB View Post
    Geesh Marko...keep it simple!

    head=1, tail=2
    111 n/a
    112 *
    121 *
    122
    211 *
    212
    221
    222 n/a

    3/6
    The original post specifies that he is required to use the rule $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$. (S)he was keeping it simple.
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  7. #7
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    Re: coin tossed

    Quote Originally Posted by SlipEternal View Post
    The original post specifies that he is required to use the rule.
    Is that the way you took it?
    Actual wording shown: "i am using the conditional probability rule".
    I took it as HIS decision.
    A thousand apologies of which you may have one
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    Re: coin tossed

    Quote Originally Posted by SlipEternal View Post
    The original post specifies that he is required to use the rule $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$. (S)he was keeping it simple.
    Here is a YouTube I have often used.
    Note the form of conditional probability is more than $\dfrac{P(A\cap B)}{P(B)}$.

    In this particular problem I bet the point is that the given is meant to be used to limit the space.
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