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Math Help - Probability question

  1. #1
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    Probability question

    I'm studying for a test and having trouble with this review question. Would you use binomial distribution on this? So convert your 2% to a faction 1/50 and then finish the binomial distribution. I see how that would work for a) but not sure about b) and c). Any guidance would be appreciated thanks.

    A light bulb manufacture claims that 2% of their light bulbs are thrown out due to defects. If you sample 50 light bulbs, what is the probability of finding :
    a) 3 defective?
    b) 2 or less defective?
    c) More than 3 defective?
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  2. #2
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    Quote Originally Posted by action259 View Post
    I'm studying for a test and having trouble with this review question. Would you use binomial distribution on this? So convert your 2% to a faction 1/50 and then finish the binomial distribution. I see how that would work for a) but not sure about b) and c). Any guidance would be appreciated thanks.

    A light bulb manufacture claims that 2% of their light bulbs are thrown out due to defects. If you sample 50 light bulbs, what is the probability of finding :
    a) 3 defective?
    b) 2 or less defective?
    c) More than 3 defective?
    I am just learning the binomial distribution in class now, so I will have a go at your questions.

    I would use a binomial for this question, definitely. As you have a fixed number of trials, and a fixed probability of success/failure

    Let X = the number of defective light bulbs in a sample of 50,

    than X~B(50, 0.02)

    than just use the formula, as you cant use the tables.  P(X=x) =\displaystyle \binom{n}{x} p^{x} (1-p)^{n-x}

    P(X=3) ?
     P(X \leq2) ?
    P(X>3)?
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  3. #3
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    Hello, action259!

    A light bulb manufacture claims that 2% of their light bulbs are defective.
    We have: . \begin{Bmatrix}P(\text{d{e}f.}) &=& \frac{1}{50} \\ \\[-3mm] P(\text{good}) &=& \frac{49}{50} \end{Bmatrix}


    If you sample 50 light bulbs, what is the probability of finding:

    a) 3 defective?
    P(\text{3 d{e}f}) \;=\;{50\choose3}\left(\frac{1}{50}\right)^3\left(  \frac{49}{50}\right)^{47}


    b) 2 or less defective?
    "2 or less" means: .0 def or 1 def or 2 def.

    . . P(\text{0 d{e}f}) \;=\;{50\choose0}\left(\frac{1}{50}\right)^0\left(  \frac{49}{50}\right)^{50}

    . . P(\text{1 d{e}f}) \;=\;{50\choose1}\left(\frac{1}{50}\right)\left(\f  rac{49}{50}\right)^{49}

    . . P(\text{2 d{e}f}) \;=\;{50\choose2}\left(\frac{1}{50}\right)^2\left(  \frac{49}{50}\right)^{48}


    The answer is the sum of the three probabilities.

    . . P(\text{2 or less}) \;=\;P(\text{0 d{e}f}) + P(\text{1 d{e}f}) + P(\text{2 d{e}f})



    c) More than 3 defective?
    The opposite of "more than 3" is "3 or less."


    P(\text{3 or less}) \:=\:\underbrace{P(\text{0 d{e}f}) + P(\text{1 d{e}f}) + P(\text{2 d{e}f})}_{\text{We found this in part (b)}} \:+\: P(\text{3 d{e}f})

    So we need: . P(\text{3 d{e}f}) \:=\:{50\choose3}\left(\frac{1}{50}\right)^3\left(  \frac{49}{50}\right)^{47}


    Add it to the answer in part (b),

    . . then subtract the sum from 1 (one).

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