54% of U. S. adults have very little confidence in newspapers. You randomly select 10 U. S. adults. Find the probability that the number of U. S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c)
less than four.
(a) P(5) =
(Round to three decimal places as needed.)
(b) P(x26) =
(Round to three decimal places as needed.)
(c) P(x<4) =
(Round to three decimal places as needed.)

Part B is not clear and the clear one is;

P(X ≥ 6)

A) 0.238

B) 0.478

C) 0.114

Step-by-step explanation:

To solve this, we will make use of binomial probability formula;

P(X = x) = nCx × p^(x)•(1 - p) ^(n - x)

A) 54% of U.S. adults have very little confidence in newspapers. Thus;

p = 0.54

10 random adults are selected. Thus;

P(X = 5) = 10C5 × 0.54^(5) × (1 - 0.54)^(10 - 5)

P(X = 5) = 0.238

B) P(X ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10)

From online binomial probability calculator, we have;

P(X ≥ 6) = 0.2331 + 0.1564 + 0.0688 + 0.01796 + 0.0021 = 0.47836 ≈ 0.478

C) P(x<4) = P(3) + P(2) + P(1) + P(0)

Again with online binomial probability calculations, we have;

P(x<4) = 0.1141 ≈ 0.114

the answer to this question is yes