Question

43% of U.5. adults have very litie confidence in newspapers. You randomly select to U.S. aduls. Find the probabily ifat the number of U.5. stuts who have vary ints confderee in newipapert is (a) exacty five, (b) at least soz, and (c) less than four. (a) P(5)= (Round wo throe decimal places as needed.)

Answer 1

To 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, we use the binomial probability formula. The probabilities are (a) P(5)=0, (b) P(X>=6)=0, (c) P(X<4)=1.2148.

Step-by-step explanation:

To 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, we need to use the binomial probability formula. The binomial formula is given by:

P(X=k) = (n choose k) * p^k * (1-p)^(n-k)

where n is the number of trials, k is the number of successes, and p is the probability of success. In this case, n=2, p=0.43, and the desired probabilities are:

(a) P(5) = (2 choose 5) * 0.43^5 * (1-0.43)^(2-5) = 0

(b) P(X>=6) = P(X=6) + P(X=7) + P(X=8) = (2 choose 6) * 0.43^6 * (1-0.43)^(2-6) + (2 choose 7) * 0.43^7 * (1-0.43)^(2-7) + (2 choose 8) * 0.43^8 * (1-0.43)^(2-8) = 0

(c) P(X<4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = (2 choose 0) * 0.43^0 * (1-0.43)^(2-0) + (2 choose 1) * 0.43^1 * (1-0.43)^(2-1) + (2 choose 2) * 0.43^2 * (1-0.43)^(2-2) + (2 choose 3) * 0.43^3 * (1-0.43)^(2-3) = 0.3643 + 0.4869 + 0.2727 + 0.0909 = 1.2148