SOLUTION: A special deck of cards has 10 green cards , 13 blue cards , and 5 red cards . When a card is picked, the color is recorded. An experiment consists of first picking a card and then

Question 1166976: A special deck of cards has 10 green cards , 13 blue cards , and 5 red cards . When a card is picked, the color is recorded. An experiment consists of first picking a card and then tossing a coin.
a. How many elements are there in the sample space?
b. Let A be the event that a green card is picked first, followed by landing a head on the coin toss.
P(A) =
Round your answer to 4 decimal places.
c. Let B be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive?
- No, they are not Mutually Exclusive
-Yes, they are Mutually Exclusive
d. Let C be the event that a green or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive?
-No, they are not Mutually Exclusive
-Yes, they are Mutually Exclusive

Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here are the solutions for the experiment involving picking a card and then tossing a coin.
## a. How many elements are there in the sample space?
The experiment consists of two independent steps:
1. **Picking a card:** There are three possible outcomes (Green, Blue, Red).
* Total cards: $10 + 13 + 5 = 28$.
2. **Tossing a coin:** There are two possible outcomes (Head, Tail).
The total number of elements in the sample space ($S$) is the product of the outcomes of each step:
$$\text{Elements in } S = (\text{Card Outcomes}) \times (\text{Coin Outcomes}) = 3 \times 2 = \mathbf{6}$$
The actual sample space is: $\{ (G, H), (G, T), (B, H), (B, T), (R, H), (R, T) \}$.
---
## b. Let A be the event that a green card is picked first, followed by landing a head on the coin toss.
The event $A$ is the outcome $(G, H)$. Since the two events are independent, $P(A) = P(\text{Green}) \times P(\text{Head})$.
1. **Probability of Green Card:**
$$P(G) = \frac{\text{Number of Green Cards}}{\text{Total Cards}} = \frac{10}{28}$$
2. **Probability of Head:**
$$P(H) = \frac{1}{2} = 0.5$$
3. **Probability of Event A:**
$$P(A) = P(G) \times P(H) = \frac{10}{28} \times 0.5 = \frac{5}{28}$$
$$P(A) \approx 0.178571...$$
Rounding to 4 decimal places:
$$P(A) = \mathbf{0.1786}$$
---
## c. Let B be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive?
* Event A: A green card is picked, followed by a Head. $A = \{(G, H)\}$
* Event B: A red or blue card is picked, followed by a Head. $B = \{(R, H), (B, H)\}$
Two events are **mutually exclusive** if they cannot occur at the same time, meaning their intersection is empty ($A \cap B = \emptyset$).
Since the card pick must be either Green (for A) or Red/Blue (for B), there are no outcomes they share.
**Answer:** **Yes, they are Mutually Exclusive**
---
## d. Let C be the event that a green or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive?
* Event A: A green card is picked, followed by a Head. $A = \{(G, H)\}$
* Event C: A green or blue card is picked, followed by a Head. $C = \{(G, H), (B, H)\}$
The intersection of $A$ and $C$ is the outcome $(G, H)$. Since $A \cap C = \{(G, H)\}$, the intersection is not empty, and $P(A \cap C) > 0$.
**Answer:** **No, they are not Mutually Exclusive**
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