Question

a special deck of cards has ten cards. four are green (g), two are blue (b), and four are red (r). when a card is picked, the color of it is recorded. an experiment consists of first picking a card and then tossing a coin, which lands on heads (h) or tails (t). part (a) part (b) part (c) part (d) let c be the event that a red or blue is picked, followed by landing a head on the coin toss. are the events a and c mutually exclusive? explain your answer in one to three complete sentences, including numerical justification (enter your probability as a fraction ) a and c select mutually exclusive because they select happen at the same time thus, p(a and c) =

Explanation:

To determine if events \( A \) (though \( A \) isn't defined here, we assume \( A \) might be a similar color - coin event, say green and head for example, but we'll work with \( C \) and the logic of mutual exclusivity) and \( C \) (red or blue picked, then head) are mutually exclusive, we recall: Two events are mutually exclusive if they cannot occur at the same time, i.e., \( P(A\cap C) = 0 \).

First, we need to know the definition of event \( A \). Since the problem is about a card - coin experiment, let's assume (since the problem is about mutual exclusivity with \( C \)) that \( A \) is an event like "green card picked and head on coin toss" (a common complementary - like event to \( C \) in such problems).

Step 1: Analyze the sample space

The deck has 10 cards: 4 green (\( G \)), 2 blue (\( B \)), 4 red (\( R \)). The coin has 2 outcomes: \( H \) (head) and \( T \) (tail). So the total number of outcomes in the sample space is \( 10\times2=20 \).

Step 2: Analyze event \( C \)

Event \( C \): red or blue card picked, then head. Number of red or blue cards: \( 4 + 2=6 \). Number of outcomes for \( C \): \( 6\times1 = 6 \) (since coin is head).

Step 3: Analyze the intersection \( A\cap C \)

If \( A \) is "green card picked and head" and \( C \) is "red or blue card picked and head", then a card can't be both green and (red or blue) at the same time. So the intersection of the card - picking parts (green vs red/blue) is empty. Then the intersection of \( A \) and \( C \) (which are combinations of card - picking and coin - tossing) will also be empty because the card - picking events are mutually exclusive.

Mathematically, the number of outcomes in \( A\cap C \) is 0 (since a card can't be green and red/blue simultaneously, and the coin outcome is head for both, but the card part is impossible). So \( P(A\cap C)=\frac{\text{Number of outcomes in }A\cap C}{\text{Total number of outcomes}}=\frac{0}{20} = 0 \).

Since \( P(A\cap C) = 0 \), the events \( A \) and \( C \) are mutually exclusive because they cannot happen at the same time (the card can't be green and red/blue at the same time, so the combined events can't occur together).

If we assume \( A \) is "green and head" (a typical event to contrast with \( C \)):

• The first "Select" dropdown: "are" (because \( P(A\cap C) = 0 \), so they are mutually exclusive)
• The second "Select" dropdown: "cannot" (because they can't happen at the same time)
• \( P(A\cap C)=\frac{0}{20}=0=\frac{0}{1} \) (but as a fraction, since there are 0 favorable outcomes out of 20 total, \( P(A\cap C) = 0=\frac{0}{1} \), but more accurately, since the card - picking events are mutually exclusive, the probability is 0. If we write it as a fraction, \( \frac{0}{20}=\frac{0}{1} \), but the simplest form is \( 0=\frac{0}{1} \), but usually we can write \( \frac{0}{20} \) or just 0. However, following the problem's instruction to enter as a fraction, we can write \( \frac{0}{20} \) (or simplify to \( \frac{0}{1} \), but 0 is also a fraction with denominator 1).

Answer:

A and C \(\boldsymbol{\text{are}}\) mutually exclusive because they \(\boldsymbol{\text{cannot}}\) happen at the same time. Thus, \( P(A \text{ and } C) = \boldsymbol{\frac{0}{20}}\) (or simplified \(\boldsymbol{0}\) or \(\boldsymbol{\frac{0}{1}}\))