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.

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