Answer :
To find the probability of getting exactly two tails when three fair coins are tossed, we can use the binomial probability formula. A binomial experiment is characterized by:
1. A fixed number of trials (n).
2. Each trial has two possible outcomes: success or failure.
3. The probability of success (p) is the same for each trial.
4. The trials are independent.
In our scenario:
- Each trial corresponds to one coin toss.
- Success (tail) and failure (head) each have a probability of 0.5.
- There are 3 trials (since three coins are tossed).
We can denote:
- [tex]\( n = 3 \)[/tex] (number of coin tosses)
- [tex]\( k = 2 \)[/tex] (number of tails desired, which we consider as successes)
- [tex]\( p = 0.5 \)[/tex] (probability of getting tails in a single toss)
- [tex]\( q = 1 - p = 0.5 \)[/tex] (probability of getting heads in a single toss)
The binomial probability formula is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k} \][/tex]
Where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, calculated as [tex]\( \frac{n!}{k!(n-k)!} \)[/tex].
Substituting our values into the formula:
[tex]\[ P(X = 2) = \binom{3}{2} \cdot (0.5)^2 \cdot (0.5)^{3-2} \][/tex]
First, calculate the binomial coefficient:
[tex]\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \][/tex]
Then, calculate the probability term:
[tex]\[ (0.5)^2 = 0.25 \quad \text{and} \quad (0.5)^{3-2} = (0.5)^1 = 0.5 \][/tex]
Now substitute these values back into the binomial probability formula:
[tex]\[ P(X = 2) = 3 \cdot 0.25 \cdot 0.5 = 3 \cdot 0.125 = 0.375 \][/tex]
So, the probability of getting exactly two tails when three fair coins are tossed is
[tex]\[ P(X = 2) = 0.375 \][/tex]
In fractional form, we can say:
[tex]\[ 0.375 = \frac{3}{8} \][/tex]
Thus, the correct answer is:
[tex]\(\frac{3}{8}\)[/tex]
1. A fixed number of trials (n).
2. Each trial has two possible outcomes: success or failure.
3. The probability of success (p) is the same for each trial.
4. The trials are independent.
In our scenario:
- Each trial corresponds to one coin toss.
- Success (tail) and failure (head) each have a probability of 0.5.
- There are 3 trials (since three coins are tossed).
We can denote:
- [tex]\( n = 3 \)[/tex] (number of coin tosses)
- [tex]\( k = 2 \)[/tex] (number of tails desired, which we consider as successes)
- [tex]\( p = 0.5 \)[/tex] (probability of getting tails in a single toss)
- [tex]\( q = 1 - p = 0.5 \)[/tex] (probability of getting heads in a single toss)
The binomial probability formula is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k} \][/tex]
Where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, calculated as [tex]\( \frac{n!}{k!(n-k)!} \)[/tex].
Substituting our values into the formula:
[tex]\[ P(X = 2) = \binom{3}{2} \cdot (0.5)^2 \cdot (0.5)^{3-2} \][/tex]
First, calculate the binomial coefficient:
[tex]\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \][/tex]
Then, calculate the probability term:
[tex]\[ (0.5)^2 = 0.25 \quad \text{and} \quad (0.5)^{3-2} = (0.5)^1 = 0.5 \][/tex]
Now substitute these values back into the binomial probability formula:
[tex]\[ P(X = 2) = 3 \cdot 0.25 \cdot 0.5 = 3 \cdot 0.125 = 0.375 \][/tex]
So, the probability of getting exactly two tails when three fair coins are tossed is
[tex]\[ P(X = 2) = 0.375 \][/tex]
In fractional form, we can say:
[tex]\[ 0.375 = \frac{3}{8} \][/tex]
Thus, the correct answer is:
[tex]\(\frac{3}{8}\)[/tex]