Suppose Sarah a student likes a local restaurant, she always orders from there one dish on both Fri and Sat. On Fri, the restaurant offers 7 different dishes on the menu among whic... Suppose Sarah a student likes a local restaurant, she always orders from there one dish on both Fri and Sat. On Fri, the restaurant offers 7 different dishes on the menu among which the first five are $10 each and the next two are $15 each. On Sat, the restaurant also offers 7 dishes where the first $10 item from Friday item is replaced with a $20 item, and the remaining six dishes are the same as those on Fri. Suppose Sarah orders independently each time regardless of the day and she doesn't order on other days. a. Write down the sample space of unique pair of orders if Sarah orders one dish both on Fri and Sat. b. Now let A be the event that Sarah orders a $10 dish in the week, B be the event that Sarah orders a $15 dish in the week, and C be the event that Sarah orders a $20 dish in the week. i. Are A and B independent? Justify your answer with calculations. ii. Are A and C independent? Justify your answer with calculations. Read more

Steps to Solve

1. Define the sample space

The sample space represents all possible unique orders Sarah can make. Since she orders one dish each on Friday and Saturday, and the dishes are spaghetti ($S$), burger ($B$), and pizza ($P$), the sample space $\Omega$ consists of all possible pairs of dishes:

$\Omega = {(S, S), (S, B), (S, P), (B, S), (B, B), (B, P), (P, S), (P, B), (P, P)}$

2. Define event A

Event $A$ is "the price of the dish Sarah orders on Friday is less than or equal to $10". The dishes that satisfy this condition are spaghetti ($S$) and burger ($B$). So, event $A$ consists of the following outcomes:

$A = {(S, S), (S, B), (S, P), (B, S), (B, B), (B, P)}$

3. Define event B

Event $B$ is "the price of the dish Sarah orders on Saturday is greater than $10". Only pizza ($P$) satisfies this, so event $B$ consists of the following outcomes:

$B = {(S, P), (B, P), (P, P)}$

4. Calculate $P(A)$

Since each outcome in the sample space is equally likely, the probability of event $A$ is the number of outcomes in $A$ divided by the total number of outcomes in $\Omega$.

$P(A) = \frac{|A|}{|\Omega|} = \frac{6}{9} = \frac{2}{3}$

5. Calculate $P(B)$

Similarly, the probability of event $B$ is the number of outcomes in $B$ divided by the total number of outcomes in $\Omega$.

$P(B) = \frac{|B|}{|\Omega|} = \frac{3}{9} = \frac{1}{3}$

6. Define event $A \cap B$

The event $A \cap B$ represents the intersection of events $A$ and $B$, meaning both events $A$ and $B$ occur. The outcomes in $A \cap B$ are those that are in both $A$ and $B$.

$A \cap B = {(S, P), (B, P)}$

7. Calculate $P(A \cap B)$

The probability of $A \cap B$ is the number of outcomes in $A \cap B$ divided by the total number of outcomes in $\Omega$.

$P(A \cap B) = \frac{|A \cap B|}{|\Omega|} = \frac{2}{9}$

8. Check for independence

Two events $A$ and $B$ are independent if $P(A \cap B) = P(A) \cdot P(B)$. Let's check if this condition holds:

$P(A) \cdot P(B) = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$

Since $P(A \cap B) = \frac{2}{9}$ and $P(A) \cdot P(B) = \frac{2}{9}$, $P(A \cap B) = P(A) \cdot P(B)$, which means that events $A$ and $B$ are independent.