Question

consider the following events
a. the burrito is a chicken burrito.
b. the burrito is a carne asada burrito.
c. the customer requested black beans.
d. the customer requested pinto beans.
which two events are independent?
a and c
a and d
b and c
b and d

Explanation:

Step1: Recall the formula for independent events

Two events \(A\) and \(C\) are independent if \(P(A\cap C)=P(A)\times P(C)\)

Step2: Calculate \(P(A)\), \(P(C)\) and \(P(A\cap C)\)

• \(P(A)=\frac{n(A)}{n(S)}=\frac{83}{240}\)
• \(P(C)=\frac{n(C)}{n(S)}=\frac{45}{240}\)
• \(P(A\cap C)=\frac{37}{240}\)
• \(P(A)\times P(C)=\frac{83}{240}\times\frac{45}{240}=\frac{83\times45}{240\times240}=\frac{3735}{57600}\approx0.065\)
• \(\frac{37}{240}\approx0.154\)

Step3: Calculate \(P(A)\), \(P(D)\) and \(P(A\cap D)\)

• \(P(A)=\frac{83}{240}\)
• \(P(D)=\frac{72}{240}\)
• \(P(A\cap D)=\frac{30}{240}\)
• \(P(A)\times P(D)=\frac{83}{240}\times\frac{72}{240}=\frac{83\times72}{240\times240}=\frac{5976}{57600}=\frac{30}{240}\) (after simplification \(\frac{83\times72}{240\times240}=\frac{83\times3}{240\times10}=\frac{249}{2400}\), and \(\frac{30}{240}=\frac{300}{2400}\). Wait, no: \(\frac{83}{240}\times\frac{72}{240}=\frac{83\times72}{240\times240}=\frac{83\times3}{240\times10}=\frac{249}{2400}\), \(\frac{30}{240}=\frac{300}{2400}\). Wait, correct calculation: \(\frac{83}{240}\times\frac{72}{240}=\frac{83\times72}{240\times240}=\frac{83\times3}{240\times10}=\frac{249}{2400}\), \(\frac{30}{240}=\frac{300}{2400}\). No, actual: \(P(A)\times P(D)=\frac{83}{240}\times\frac{72}{240}=\frac{83\times72}{240\times240}=\frac{83\times3}{240\times10}=\frac{249}{2400}\), \(P(A\cap D)=\frac{30}{240}=\frac{300}{2400}\). Wait, wrong approach. Correct: \(P(A)=\frac{83}{240}\), \(P(D)=\frac{72}{240}\), \(P(A\cap D)=\frac{30}{240}\). \(P(A)\times P(D)=\frac{83\times72}{240\times240}=\frac{83\times3}{240\times10}=\frac{249}{2400}\approx0.104\), \(\frac{30}{240} = 0.125\). No. Wait, correct formula: For two - way table, \(P(A)=\frac{83}{240}\), \(P(D)=\frac{72}{240}\), \(n(A\cap D) = 30\), \(P(A\cap D)=\frac{30}{240}\). \(P(A)\times P(D)=\frac{83\times72}{240\times240}=\frac{83\times3}{240\times10}=\frac{249}{2400}\), \(\frac{30}{240}=\frac{300}{2400}\). No. Wait, another way: \(n(A) = 83\), \(n(D)=72\), \(n(A\cap D) = 30\). \(P(A)\times P(D)=\frac{83}{240}\times\frac{72}{240}\), \(P(A\cap D)=\frac{30}{240}\). Cross - multiply: \(83\times72=5976\), \(30\times240 = 7200\). Not equal.

Step4: Calculate \(P(B)\), \(P(C)\) and \(P(B\cap C)\)

• \(P(B)=\frac{80}{240}\)
• \(P(C)=\frac{45}{240}\)
• \(P(B\cap C)=\frac{5}{240}\)
• \(P(B)\times P(C)=\frac{80\times45}{240\times240}=\frac{3600}{57600}=\frac{5}{240}\)

Step5: Calculate \(P(B)\), \(P(D)\) and \(P(B\cap D)\)

• \(P(B)=\frac{80}{240}\)
• \(P(D)=\frac{72}{240}\)
• \(P(B\cap D)=\frac{24}{240}\)
• \(P(B)\times P(D)=\frac{80\times72}{240\times240}=\frac{5760}{57600}=\frac{1}{10}=\frac{24}{240}\) (after simplification \(\frac{80\times72}{240\times240}=\frac{80\times3}{240\times10}=\frac{240}{2400}=\frac{1}{10}\), \(\frac{24}{240}=\frac{1}{10}\))

Answer:

B and C