Two dice are rolled, and the events Fand T are as follows: F = {The sum of the dice is four} and T = (At least one die shows a three) Then P(F U T) is: 15/36 None of these 13/36 12/36
Two dice are rolled, and the events Fand T are as follows: F = {The sum of the dice is four} and T = (At least one die shows a three) Then P(F U T) is: 15/36 None of these 13/36 12/36
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![**Problem:**
Two dice are rolled, and the events F and T are as follows:
- F = {The sum of the dice is four}
- T = {At least one die shows a three}
Then \( P(F \cup T) \) is:
- [ ] \( \frac{15}{36} \)
- [x] None of these
- [ ] \( \frac{13}{36} \)
- [ ] \( \frac{12}{36} \)
**Explanation:**
To solve for \( P(F \cup T) \), we need to find the probability that either event F occurs, event T occurs, or both occur.
- **Event F:** The possible outcomes where the sum of the dice is four are:
- (1, 3)
- (2, 2)
- (3, 1)
So, there are 3 outcomes for event F.
- **Event T:** The possible outcomes where at least one die shows a three are:
- (1, 3)
- (2, 3)
- (3, 1)
- (3, 2)
- (3, 3)
- (3, 4)
- (3, 5)
- (3, 6)
- (4, 3)
- (5, 3)
- (6, 3)
So, there are 11 outcomes for event T.
- **Event \( F \cap T \):** The outcomes common to both F and T are:
- (1, 3)
- (3, 1)
So, there are 2 outcomes for \( F \cap T \).
Using the formula for the union of two events:
\[ P(F \cup T) = P(F) + P(T) - P(F \cap T) \]
\[ P(F) = \frac{3}{36}, \quad P(T) = \frac{11}{36}, \quad P(F \cap T) = \frac{2}{36} \]
\[ P(F \cup T) = \frac{3}{36} + \frac{11}{36} - \frac{2}{36} = \frac{12}{36} \]
Thus, the correct probability that either event F or T, or both,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9a59ff3-32cb-4a72-806c-172db4d1cb49%2Fd16e04bf-200f-44c2-80a4-f38fbac9aacf%2Fogs3v_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Two dice are rolled, and the events F and T are as follows:
- F = {The sum of the dice is four}
- T = {At least one die shows a three}
Then \( P(F \cup T) \) is:
- [ ] \( \frac{15}{36} \)
- [x] None of these
- [ ] \( \frac{13}{36} \)
- [ ] \( \frac{12}{36} \)
**Explanation:**
To solve for \( P(F \cup T) \), we need to find the probability that either event F occurs, event T occurs, or both occur.
- **Event F:** The possible outcomes where the sum of the dice is four are:
- (1, 3)
- (2, 2)
- (3, 1)
So, there are 3 outcomes for event F.
- **Event T:** The possible outcomes where at least one die shows a three are:
- (1, 3)
- (2, 3)
- (3, 1)
- (3, 2)
- (3, 3)
- (3, 4)
- (3, 5)
- (3, 6)
- (4, 3)
- (5, 3)
- (6, 3)
So, there are 11 outcomes for event T.
- **Event \( F \cap T \):** The outcomes common to both F and T are:
- (1, 3)
- (3, 1)
So, there are 2 outcomes for \( F \cap T \).
Using the formula for the union of two events:
\[ P(F \cup T) = P(F) + P(T) - P(F \cap T) \]
\[ P(F) = \frac{3}{36}, \quad P(T) = \frac{11}{36}, \quad P(F \cap T) = \frac{2}{36} \]
\[ P(F \cup T) = \frac{3}{36} + \frac{11}{36} - \frac{2}{36} = \frac{12}{36} \]
Thus, the correct probability that either event F or T, or both,
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