Question 1177710: A coin is tossed three times. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses. If the coin is unbalanced and a
head has a 40% chance of occurring, find the joint probability distribution of W and Z.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**1. Define Random Variables**
* **Z:** Number of heads on the first toss (Z can be 0 or 1)
* **W:** Total number of heads in the last two tosses (W can be 0, 1, or 2)
**2. Probabilities**
* Probability of a head (H): P(H) = 0.4
* Probability of a tail (T): P(T) = 1 - 0.4 = 0.6
**3. Possible Outcomes and Corresponding W and Z Values**
Let's list all possible outcomes of three coin tosses and their corresponding W and Z values:
| Outcome | Z (Heads on 1st Toss) | W (Heads on 2nd & 3rd Tosses) | Probability |
|---|---|---|---|
| TTT | 0 | 0 | (0.6)(0.6)(0.6) = 0.216 |
| TTH | 0 | 1 | (0.6)(0.6)(0.4) = 0.144 |
| THT | 0 | 1 | (0.6)(0.4)(0.6) = 0.144 |
| THH | 0 | 2 | (0.6)(0.4)(0.4) = 0.096 |
| HTT | 1 | 0 | (0.4)(0.6)(0.6) = 0.144 |
| HTH | 1 | 1 | (0.4)(0.6)(0.4) = 0.096 |
| HHT | 1 | 1 | (0.4)(0.4)(0.6) = 0.096 |
| HHH | 1 | 2 | (0.4)(0.4)(0.4) = 0.064 |
**4. Joint Probability Distribution**
Now, we create the joint probability distribution table:
| Z\W | 0 | 1 | 2 |
|---|---|---|---|
| 0 | 0.216 | 0.288 (0.144 + 0.144) | 0.096 |
| 1 | 0.144 | 0.192 (0.096 + 0.096) | 0.064 |
**Verification**
Let's verify that the sum of all probabilities is 1:
0.216 + 0.288 + 0.096 + 0.144 + 0.192 + 0.064 = 1.000
**Therefore, the joint probability distribution of W and Z is:**
| Z\W | 0 | 1 | 2 |
|---|---|---|---|
| 0 | 0.216 | 0.288 | 0.096 |
| 1 | 0.144 | 0.192 | 0.064 |
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