Probability_Math

Question

​58% 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(x≥6)=(Round to three decimal places as​ needed.)

​(c) ​P(x​<4)=(Round to three decimal places as​ needed.

Answer 1

This is modeled with a binomial random variable:

1. There are a fixed number (10) people questioned
2. The probability of success (i.e. very little confidence in newspapers) is constant trial to trial (p=0.58)
3. The trials are independent (a reasonable assumption, since the response of 1 of the people should not affect the response of any other)
4. Each trial has only two outcomes - little confidence in newspapers, or not
5. X counts the number of successes in the 10 trials

The probability mass function for the binomial random variable is given by

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

a) P(X=5) = (10choose5)*(0.58)^5*(0.42)^5 ≈ 0.216

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

Each of these can be computed separately, and then added together

c) P(X<4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)

Same idea as b) above.

Alternatively, if you have a calculator with the binomial's cumulative distribution function, you can thus that for b) and c) directly.

Recall that the binomcdf(x) = P(X ≤ x) - note the direction and type of the inequality.

So, for b), we would have P(X ≥6) = 1-P(X<6) = 1-binomcdf(5) (we use 5 here because we want the probability of X being strictly less than 6).

For c), it would be P(X<4) = binomcdf(3)