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Math Help - Computation of probabilities in Bernoulli Trials

  1. #1
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    Computation of probabilities in Bernoulli Trials

    (1) Find C(n, x)pxqn −x for the given values of n, x, and p. (Round your answer to four decimal places.)

    n = 7, x = 2, p = 1/3
    (2) Use the formula C(n, x)pxqnx to determine the probability of the given event. (Round your answer to four decimal places.)

    The probability of exactly three successes in six trials of a binomial experiment in which p = 1/2


    (3) Use the formula C(n, x)pxqnx to determine the probability of the given event. (Round your answer to four decimal places.)

    The probability of at least five successes in ten trials of a binomial experiment in which p = 1/2



    (4) Use the formula C(n, x)pxqnx to determine the probability of the given event. (Round your answer to four decimal places.)
    The probability of no failures in seven trials of a binomial experiment in which p = 1/3
    (5) Use the formula C(n, x)pxqnx to determine the probability of the given event. (Round your answer to four decimal places.)

    The probability of at least one failure in seven trials of a binomial experiment in which p = 1/4
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  2. #2
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    Re: Computation of probabilities in Bernoulli Trials

    Hello, foz124!

    (1) Given: . n = 7,\;x=2,\;p=\tfrac{1}{3}
    . . Find C(n,x)p^xq^{n-z} . (Round your answer to four decimal places.)
    C(7,2)\left(\tfrac{1}{3}\right)^2\left(\tfrac{2}{3  }\right)^5 \;=\;\frac{672}{2187} \;=\;0.307270233 \;\approx\;0.3073



    (2) The probability of exactly 3 successes in 6 trials: . p = \tfrac{1}{2}
    C(6,3)\left(\tfrac{1}{2}\right)^3\left(\tfrac{1}{2  }\right)^3 \;=\;\frac{20}{64} \;=\;0.3125



    (3) The probability of at least 5 successes in 10 trials: . p = \tfrac{1}{2}
    "At least 5 successes" means 5, 6, 7, 8, 9, or 10 successes.
    We must determine the probabilty of each case . . . then add them.

    \begin{array}{cccccc}\text{5 successes:} & C(10,5)(\frac12)^5(\frac12)^5 &=& \frac{252}{1024} \\ \\[-4mm] \text{6 successes:} & C(10,6)(\frac12)^6(\frac12)^4 &=& \frac{210}{1024} \\ \\[-4mm] \text{7 successes:} & C(10,7)(\frac12)^7(\frac12)^3 &=& \frac{120}{1024} \\ \\[-4mm] \text{8 successes:} & C(10,8)(\frac12)^8(\frac12)^2 &=& \frac{45}{1024} \\ \\[-4mm] \text{9 successes:} & C(10,9)(\frac12)^9(\frac12)^1 &=& \frac{10}{1024} \\ \\[-4mm] \text{10 successes:} & C(10,10)(\frac12)^{10}(\frac12)^0 &=& \frac{1}{1024} \end{array}

    Total: . \frac{638}{1024} \;=\;0.623046875 \;\approx\;0.6230




    (4) The probability of no failures in 7 trials: . p =  \tfrac{1}{3}
    "No failures" means "7 successes."

    C(7,7)(\tfrac13)^7(\tfrac23)^0 \;=\;\frac{1}{2187} \;=\;0.000457247 \;\approx\;0.0005




    (5) The probability of at least one failure in 7 trials: . p = \tfrac{1}{4}
    The opposite of "at least one failure" is "no failures" (all successes).

    P(\text{7 successes}) \;=\;C(7,7)(\tfrac14)^7(\tfrac{3}{4})^0 \;=\;\frac{1}{16,384}

    P(\text{at least one failure}) \;=\;1 - \frac{1}{16,384} \;=\;0.999938965 \;\approx\;0.9999

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  3. #3
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    Re: Computation of probabilities in Bernoulli Trials

    Terrific! My book sometimes confuses me. I appreciate it!
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