Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pa

Question

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities:

(a) P(all of the next three vehicles inspected pass)
(b) P(at least one of the next three inspected fails)
(c) P(exactly one of the next three inspected passes)
(d) P(at most one of the next three vehicles inspected passes)
(e) Given that at least one of the next three vehicles p

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Rylee 1 week 2022-01-12T05:20:45+00:00 2 Answers 0 views 0

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    0
    2022-01-12T05:21:48+00:00

    Answer:

    (a) 0.343

    (b) 0.657

    (c) 0.189

    (d) 0.216

    (e) 0.353

    Step-by-step explanation:

    Let P(a vehicle passing the test) = p

                            p = \frac{70}{100} = 0.7  

    Let P(a vehicle not passing the test) = q

                             q = 1 – p

                             q = 1 – 0.7 = 0.3

    (a) P(all of the next three vehicles inspected pass) = P(ppp)

                               = 0.7 × 0.7 × 0.7

                               = 0.343

    (b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

          = (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

          = 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

          = 0.657

    (c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

                     =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

                     = 0.063 + 0.063 + 0.063

                     = 0.189

    (d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

                     =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

                     = 0.063 + 0.063 + 0.063 + 0.027

                     = 0.216

    (e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

    P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

    =  (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

    = 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

                      = 0.973  

    With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

                             = \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}

                             = \frac{0.343}{0.973}

                             = 0.353

    0
    2022-01-12T05:22:40+00:00

    Answer:

    a) The probability that all of the next three vehicles inspected pass is 0.343

    b) The probability that at least one of the next three inspected fails is 0.657

    c) The probability that exactly one of the next three vehicles inspected passes is 0.189

    d) The probability of at most one of the next three vehicles inspected passes is 0.216

    e) The probability that they all pass given that at least one of the next three vehicle pass is 0.343

    Step-by-step explanation:

    This is a binomial distribution problem.

    Probability of success which is passing (p) = 70% = \frac{7}{10}

    Probability of failing (q) = 1 – p = 1 - \frac{7}{10}  = \frac{3}{10}

    Based on the options,  number of trial (n) = 3

    The formula for binomial distribution problem is given as:

    P(X=x) = (nCx) p^{x} q^{n-x}

    nCx means n combination x

    a) P(all of the next three vehicles inspected pass)

    P(X=x) = (nCx) p^{x} q^{n-x}\\P(X=3) = (3C3) 0.7^{3} 0.3^{3-3}\\P(X=3) = (\frac{3!}{3!(3-3)!} ) 0.7^{3} 0.3^{3-3}\\P(X=3) = (\frac{3!}{3!0!} ) 0.7^{3} 0.3^{0}\\P(X=3) = 1 * 0.7^{3}  * 1\\P(X=3) = 1 * 0.7^{3}  * 1\\P(X=3) = 1 * 0.343 * 1\\P(X=3) = 0.343

    b) P(at least one of the next three inspected fails)

    Probability that at least one of the vehicle means that more that not less than one vehicle fail. So it could be one fail, two fail or three fail which means probability of exactly two pass or less than two pass.

    P(X=x) = (nCx) p^{x} q^{n-x}\\P(X\leq 2) = P(X=0) + P(X=1) + P(X=2)\\But P(X=1) = 0.189 \\P(X=0) = 0.027\\P(X=2) = 0.441\\P(X\leq 2) = 0.027 + 0.189 + 0.441\\P(X\leq 2) = 0.657

    (c) P(exactly one of the next three inspected passes)

    P(X=x) = (nCx) p^{x} q^{n-x}\\P(X=1) = (3C1) 0.7^{1} 0.3^{3-1}\\P(X=1) = (\frac{3!}{1!(3-1)!} ) 0.7^{1} 0.3^{3-1}\\P(X=1) = (\frac{3!}{1!2!} ) 0.7^{1} 0.3^{2}\\P(X=1) = (\frac{6}{1*2} ) 0.7^{1} 0.3^{2}\\P(X=1) = (\frac{6}{2} ) 0.7^{1} 0.3^{2}\\P(X=1) = 3* 0.7^{1} 0.3^{2}\\P(X=1) = 3 * 0.7 * 0.09\\P(X=1) = 0.189

    (d) P(at most one of the next three vehicles inspected passes)

    At most one means not more than one vehicles pass, that is number of vehicles less than 1

    P(X=x) = (nCx) p^{x} q^{n-x}\\P(X\leq 1) = P(X=0) + P(X=1)\\But P(X=1) = 0.189 \\P(X=0) = (3C0) 0.7^{0} 0.3^{3-0}\\P(X=0) = (\frac{3!}{0!(3-0)!} ) 0.7^{0} 0.3^{3-0}\\P(X=0) = (\frac{3!}{0!3!} ) 0.7^{0} 0.3^{3}\\P(X=0) = (\frac{6}{6} ) 0.7^{0} 0.3^{3}\\P(X=0) = 1 * 0.7^{0} 0.3^{3}\\P(X=0) = 1 * 1 * 0.027\\P(X=0) = 0.027\\\\P(X\leq 1) = P(X=0) + P(X=1)\\But P(X=1) = 0.189\\P(X=0) = 0.027\\P(X\leq 1) = 0.027 + 0.189\\P(X\leq 1) = 0.216

    (e) Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

    Formula for conditional probability is given as:

    P(B|A) = \frac{P(A n B)}{P(A)}

    where

    B is the probability that they all passes

    A is the given probability that at least one of the next three vehicle passes

    We need to find the probability for A.

    P(X=x) = (nCx) p^{x} q^{n-x}\\P(X\geq 1) = P(X=1) + P(X=2) + P(X =3)\\But\\P(X=1) = 0.441\\P(X=3) = 0.343\\P(X=2) = ?\\P(X=2) = (\frac{3!}{2!(3-2)!} ) 0.7^{2} 0.3^{3-2}\\P(X=2) = (\frac{3!}{2!1!} ) 0.7^{2} 0.3^{1}\\P(X=2) = (\frac{6}{2*1} ) 0.7^{2} 0.3^{1}\\P(X=2) = (\frac{6}{2} ) 0.7^{2} 0.3^{1}\\P(X=2) = 3 * 0.49 * 0.3\\P(X=2) = 0.441\\P(X\geq 1) = P(X=1) + P(X=2) + P(X =3)\\P(X\geq 1) = 0.189 + 0.441 + 0.343\\P(X\geq 1) =0.973

    Probability that they all pass = 0.343

    Probability that at least one pass = 0.973

    P(B|A) = \frac{P(A n B)}{P(A)}\\P(B|A) = \frac{0.343 * 0.973}{0.973} \\P(B|A) = \frac{0.333739}{0.973} \\P(B|A) = 0.343

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