Social Science According to the 2012 General Social Sur-vey, the probability that a U.S. resident has a great deal of confi-dence in medicine is .389 and the probability that a U.S. residenthas a great deal of confidence in major companies is .166. Theprobability of having a great deal of confidence in both medicineand major companies is .111. Find the probability of thedescribed events. (Data from: www3.norc.org/gss+website.)(a) Not having a great deal of confidence in medicine and hav-ing a great deal of confidence in major companies.(b) Having a great deal of confidence in medicine or not hav-ing a great deal of confidence in major companies.(c) Not having a great deal of confidence in medicine and nothaving a great deal of confidence in major companies.

Name: Social Science According to the 2012 General Social Sur-vey, the prob
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Description: Social Science According to the 2012 General Social Sur-vey, the probability that a U.S. resident has a great deal of confi-dence in medicine is .389 and the probability that a U.S. residenthas a great deal of confidence in major companies is .166.

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Social Science According to the 2012 General Social Sur-
vey, the probability that a U.S. resident has a great deal of confi-
dence in medicine is .389 and the probability that a U.S. resident
has a great deal of confidence in major companies is .166. The
probability of having a great deal of confidence in both medicine
and major companies is .111. Find the probability of the
described events. (Data from: www3.norc.org/gss+website.)
(a) Not having a great deal of confidence in medicine and hav-
ing a great deal of confidence in major companies.
(b) Having a great deal of confidence in medicine or not hav-
ing a great deal of confidence in major companies.
(c) Not having a great deal of confidence in medicine and not
having a great deal of confidence in major companies.

Expert Solution

Step 1

Let A be the event that a U.S. resident has a great deal of confidence in medicine.

Let B be the event that a U.S. resident has a great deal of confidence in major companies.

Given that, $P (A) = 0.389, P (B) = 0.166, P (A \cap B) = 0.111$

a) Here, we have to find $P (A^{c} \cap B)$

$P (A^{c} \cap B) = P (B) - P (A \cap B) P (A^{c} \cap B) = 0.166 - 0.111 P (A^{c} \cap B) = 0.055$

Hence, the probability that not having a great deal of confidence in medicine and having a great deal of confidence in major companies is 0.055.

b)Here, we have to find $P (A \cup B^{c})$

$P (A \cup B^{c}) = P (A) + P (B^{c}) - P (A \cap B^{c}) Thus, P (B^{c}) = 1 - P (B) = 1 - 0.166 = 0.834 P (A \cap B^{c}) = P (A) - P (A \cap B) = 0.389 - 0.111 = 0.278 Hence, P (A \cup B^{c}) = P (A) + P (B^{c}) - P (A \cap B^{c}) P (A \cup B^{c}) = 0.389 + 0.834 - 0.278 P (A \cup B^{c}) = 0.945$

Hence, the probability that having a great deal of confidence in medicine or not having a great deal of confidence in major companies is 0.945.

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