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
