Let's start by listing all the possible outcomes when three fair coins are tossed at the same time. Each coin has two potential outcomes: heads (H) or tails (T).
The possible outcomes are:
1. HHH
2. HHT
3. HTH
4. HTT
5. THH
6. THT
7. TTH
8. TTT
There are a total of [tex]\( 2^3 = 8 \)[/tex] possible outcomes. We will use these outcomes to answer the sub-questions.
### a) Probability of obtaining three heads (HHH)
We look for the outcome where all three coins show heads. There is only one such outcome: [tex]\( \{HHH\} \)[/tex].
Therefore, the probability [tex]\( P(\text{three heads}) \)[/tex] is:
[tex]\[
P(\text{three heads}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{1}{8}
\][/tex]
### b) Probability of obtaining two heads and one tail
We look for the outcomes where exactly two coins show heads, and one coin shows tails. These outcomes are:
- HHT
- HTH
- THH
There are 3 such outcomes.
Therefore, the probability [tex]\( P(\text{two heads and one tail}) \)[/tex] is:
[tex]\[
P(\text{two heads and one tail}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{3}{8}
\][/tex]
### c) Probability of obtaining no heads (TTT)
We look for the outcome where there are no heads. There is only one such outcome: [tex]\( \{TTT\} \)[/tex].
Therefore, the probability [tex]\( P(\text{no heads}) \)[/tex] is:
[tex]\[
P(\text{no heads}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{1}{8}
\][/tex]
### d) Probability of obtaining at least one head
We look for outcomes where there is at least one head. We can find it by either:
1. Counting all outcomes that include at least one head, or
2. Using the complement rule: subtract the probability of getting no heads from 1.
From our list of outcomes, we have:
- The outcomes without heads: [tex]\( \{TTT\} \)[/tex]
Thus, there are [tex]\( 8 - 1 = 7 \)[/tex] outcomes with at least one head.
Therefore, the probability [tex]\( P(\text{at least one head}) \)[/tex] is:
[tex]\[
P(\text{at least one head}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{7}{8}
\][/tex]
Alternatively, using the complement rule:
[tex]\[
P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \frac{1}{8} = \frac{7}{8}
\][/tex]
Both methods give the same result.
### Summary
- [tex]\( P(\text{three heads}) = \frac{1}{8} \)[/tex]
- [tex]\( P(\text{two heads and one tail}) = \frac{3}{8} \)[/tex]
- [tex]\( P(\text{no heads}) = \frac{1}{8} \)[/tex]
- [tex]\( P(\text{at least one head}) = \frac{7}{8} \)[/tex]
