Please answer all parts of the question and all work shown. Suppose a professor of probability is tired of reading the depressi

Question

Please answer all parts of the question and all work shown.

Suppose a professor of probability is tired of reading the depressing news and so he decides that he will quickly scan the first 5 headlines in the New Yorks Times and the first 5 headlines in the Boston Globe and if at most 3 of the articles in each are depressing, he will read the news that day. Further suppose that the probability of a NYTs headline being depressing is 0.6 and for the Globe the probability of a headline being depressing is 0.55.

(a) What is the probability that he will read the news the first day he tries this?

(b) In order to be “well-informed” he needs to read the news at least half the time; what is the probability that he will be well-informed after doing this for a week?

Hint: This is another problem where there are two independent parts of the random experiment. You might want to phrase it as three different random variables, all three binomial.

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Sadie 7 months 2021-10-08T15:51:44+00:00 1 Answer 0 views 0

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    2021-10-08T15:53:00+00:00

    Answer:

    a. 0.4931

    b. 0.2695

    Step-by-step explanation:

    Given

    Let BG represents Boston Globe

    NYT represents New York Times

    P(BG) = 0.55

    P(BG’) = 1 – 0.55 = 0.45

    P(NYT) = 0.6

    P(NYT’) = 1 -0.6 = 0.4

    Number of headlines = 5

    Number of depressed articles = 3 (at most)

    a.

    Let P(Read) = Probability that he reads the news the first day

    P(Read) = P(He reads BG) and P(He reads NYT)

    For the professor to read BG, then there must be at most 3 depressing news

    i.e P(0) + P(1) + P(2) + P(3)

    But P(0) + P(1) + …. + P(5) = 1 (this is the sample space)

    So,

    P(0) + P(1) + P(2) + P(3) = 1 – P(4) – P(5)

    P(4) or P(BG = 4) is given as the binomial below

    (BG + BG’)^n where n = 5, r = 4

    So, P(BG = 4) = C(5,4) * 0.55⁴ * 0.45¹

    P(BG = 5). = (BG + BG’)^n where n = 5, r = 5

    So, P(BG = 5) = C(5,5) * 0.55^5 * 0.45°

    P(0) + P(1) + P(2) + P(3)= 1 – P(BG = 4) – P(BG = 5)

    P(0) + P(1) + P(2) + P(3) = 1 – C(5,4) * 0.55⁴ * 0.45¹ – C(5,5) * 0.55^5 * 0.45°

    P(0) + P(1) + P(2) + P(3) = 0.7438

    For the professor to read NYT, then there must be at most 3 depressing news

    i.e P(0) + P(1) + P(2) + P(3)

    But P(0) + P(1) + …. + P(5) = 1 (this is the sample space)

    So,

    P(0) + P(1) + P(2) + P(3) = 1 – P(4) – P(5)

    P(4) or P(NYT = 4) is given as the binomial below

    (NYT+ NYT’)^n where n = 5, r = 4

    So, P(NYT = 4) = C(5,4) * 0.6⁴ * 0.4¹

    P(NYT = 5). = (NYT + NYT’)^n where n = 5, r = 5

    So, P(NYT = 5) = C(5,5) * 0.6^5 * 0.4°

    P(0) + P(1) + P(2) + P(3)= 1 – P(NYT = 4) – P(NYT = 5)

    P(0) + P(1) + P(2) + P(3) = 1 – C(5,4) * 0.6⁴ * 0.4¹ – C(5,5) * 0.6^5 * 0.4°

    P(0) + P(1) + P(2) + P(3) = 0.6630

    P(Read) = P(He reads BG) and P(He reads NYT)

    P(Read) = 0.7438 * 0.6630

    P(Read) = 0.4931

    b.

    Given

    n = Number of week = 7

    P(Read) = 0.4931

    R(Read’) = 1 – 0.4931 =

    He needs to read at least half the time means he reads for 4 days a week

    So,

    P(Well-informed) = (Read + Read’)^n where n = 7, r = 4

    P(Well-informed) = C(7,4) * (0.4931)⁴ * (1-0.4931)³ = 0.2695

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