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

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.

## Answers ( )

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