Answer :
To solve the problem using the theoretical probability formula, let's follow these steps:
1. List the sample space: When a fair coin is tossed two times, the possible outcomes are:
- HH (Heads on the first toss and Heads on the second toss)
- HT (Heads on the first toss and Tails on the second toss)
- TH (Tails on the first toss and Heads on the second toss)
- TT (Tails on the first toss and Tails on the second toss)
So, the sample space [tex]\( S \)[/tex] is [tex]\(\{HH, HT, TH, TT\}\)[/tex].
2. Identify the favorable outcomes: We are interested in the probability of getting the same outcome on each toss. In our sample space, the outcomes where both tosses are the same are:
- HH
- TT
Therefore, the number of favorable outcomes is 2.
3. Determine the total number of possible outcomes: There are 4 possible outcomes in the sample space.
4. Calculate the probability: The theoretical probability [tex]\( P(E) \)[/tex] of an event [tex]\( E \)[/tex] is given by the ratio of the number of favorable outcomes to the total number of possible outcomes. This can be expressed as:
[tex]\[ P(\text{same outcome on each toss}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{4} \][/tex]
5. Reduce the fraction to lowest terms:
[tex]\[ \frac{2}{4} = \frac{1}{2} \][/tex]
Thus, the probability of getting the same outcome on each toss when a fair coin is tossed two times is [tex]\(\frac{1}{2}\)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{\frac{1}{2}} \][/tex]
1. List the sample space: When a fair coin is tossed two times, the possible outcomes are:
- HH (Heads on the first toss and Heads on the second toss)
- HT (Heads on the first toss and Tails on the second toss)
- TH (Tails on the first toss and Heads on the second toss)
- TT (Tails on the first toss and Tails on the second toss)
So, the sample space [tex]\( S \)[/tex] is [tex]\(\{HH, HT, TH, TT\}\)[/tex].
2. Identify the favorable outcomes: We are interested in the probability of getting the same outcome on each toss. In our sample space, the outcomes where both tosses are the same are:
- HH
- TT
Therefore, the number of favorable outcomes is 2.
3. Determine the total number of possible outcomes: There are 4 possible outcomes in the sample space.
4. Calculate the probability: The theoretical probability [tex]\( P(E) \)[/tex] of an event [tex]\( E \)[/tex] is given by the ratio of the number of favorable outcomes to the total number of possible outcomes. This can be expressed as:
[tex]\[ P(\text{same outcome on each toss}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{4} \][/tex]
5. Reduce the fraction to lowest terms:
[tex]\[ \frac{2}{4} = \frac{1}{2} \][/tex]
Thus, the probability of getting the same outcome on each toss when a fair coin is tossed two times is [tex]\(\frac{1}{2}\)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{\frac{1}{2}} \][/tex]