Peter has two fair tetrahedral (four sided) dice. The faces on each die are labeled l, 2, 3 and 4. One die is red and another is blue. Peter throws each die once. The random variable X is the sum of the numbers on which the dice land.
a) Construct a probability distribution table for X.
b) Find the probability that X=4.
c) Given that X=6, find the probability that the red die landed on 2
X has the possible values from 2 to 8
figure the probability of each
a) To construct a probability distribution table for X, we need to determine all possible outcomes and their respective probabilities.
Let's list all the possible outcomes of throwing both dice once and finding their sums:
Outcome 1: Red die lands on 1, Blue die lands on 1. Sum = 1 + 1 = 2.
Outcome 2: Red die lands on 1, Blue die lands on 2. Sum = 1 + 2 = 3.
Outcome 3: Red die lands on 1, Blue die lands on 3. Sum = 1 + 3 = 4.
Outcome 4: Red die lands on 1, Blue die lands on 4. Sum = 1 + 4 = 5.
Outcome 5: Red die lands on 2, Blue die lands on 1. Sum = 2 + 1 = 3.
Outcome 6: Red die lands on 2, Blue die lands on 2. Sum = 2 + 2 = 4.
Outcome 7: Red die lands on 2, Blue die lands on 3. Sum = 2 + 3 = 5.
Outcome 8: Red die lands on 2, Blue die lands on 4. Sum = 2 + 4 = 6.
Outcome 9: Red die lands on 3, Blue die lands on 1. Sum = 3 + 1 = 4.
Outcome 10: Red die lands on 3, Blue die lands on 2. Sum = 3 + 2 = 5.
Outcome 11: Red die lands on 3, Blue die lands on 3. Sum = 3 + 3 = 6.
Outcome 12: Red die lands on 3, Blue die lands on 4. Sum = 3 + 4 = 7.
Outcome 13: Red die lands on 4, Blue die lands on 1. Sum = 4 + 1 = 5.
Outcome 14: Red die lands on 4, Blue die lands on 2. Sum = 4 + 2 = 6.
Outcome 15: Red die lands on 4, Blue die lands on 3. Sum = 4 + 3 = 7.
Outcome 16: Red die lands on 4, Blue die lands on 4. Sum = 4 + 4 = 8.
Now, let's calculate the probabilities for each outcome. Since both dice are fair tetrahedral dice, each face has a 1/4 chance of landing face up.
Outcome 1: Probability = (1/4) * (1/4) = 1/16
Outcome 2: Probability = (1/4) * (1/4) = 1/16
Outcome 3: Probability = (1/4) * (1/4) = 1/16
Outcome 4: Probability = (1/4) * (1/4) = 1/16
Outcome 5: Probability = (1/4) * (1/4) = 1/16
Outcome 6: Probability = (1/4) * (1/4) = 1/16
Outcome 7: Probability = (1/4) * (1/4) = 1/16
Outcome 8: Probability = (1/4) * (1/4) = 1/16
Outcome 9: Probability = (1/4) * (1/4) = 1/16
Outcome 10: Probability = (1/4) * (1/4) = 1/16
Outcome 11: Probability = (1/4) * (1/4) = 1/16
Outcome 12: Probability = (1/4) * (1/4) = 1/16
Outcome 13: Probability = (1/4) * (1/4) = 1/16
Outcome 14: Probability = (1/4) * (1/4) = 1/16
Outcome 15: Probability = (1/4) * (1/4) = 1/16
Outcome 16: Probability = (1/4) * (1/4) = 1/16
Now we can construct the probability distribution table for X:
| X | Probability |
|-------|-------------|
| 2 | 1/16 |
| 3 | 2/16 |
| 4 | 3/16 |
| 5 | 4/16 |
| 6 | 3/16 |
| 7 | 2/16 |
| 8 | 1/16 |
b) To find the probability that X=4, we look at the probability distribution table. We can see that the probability of X=4 is 3/16.
c) Given that X=6, we need to find the probability that the red die landed on 2. To calculate this, we need to consider all the outcomes where the sum is 6.
Outcomes where X=6:
- Outcome 8: Red die lands on 2, Blue die lands on 4. Sum = 2 + 4 = 6.
- Outcome 11: Red die lands on 3, Blue die lands on 3. Sum = 3 + 3 = 6.
- Outcome 14: Red die lands on 4, Blue die lands on 2. Sum = 4 + 2 = 6.
Out of these three outcomes, only one has the red die landing on 2. Therefore, the probability that the red die landed on 2, given that X=6, is 1/3.