Consider the following statements
I : An experiment succeeds twice as often as it fails. Then, the probability that in the next six trials, there will be atleast 4 successes is `31/9(2/3)^(4)`
II : The number of times must a man toss a fair coin so that the probability of having atleast one head is more than 90% is 4 or mor than 4.

Options

A. I is true
B. II is true
C. Both are true
D. Both are false

Answer

To solve the problem, we will analyze each statement one by one.

### Statement I:
**An experiment succeeds twice as often as it fails. Then, the probability that in the next six trials, there will be at least 4 successes is \( \frac{31}{9} \left( \frac{2}{3} \right)^{4} \).**

1. **Define the probabilities:**
Let the probability of failure be \( x \). Then the probability of success is \( 2x \). Since the total probability must equal 1, we have:
\[ x + 2x = 1 \implies 3x = 1 \implies x = \frac{1}{3} \]
Therefore, the probability of failure \( P(F) = \frac{1}{3} \) and the probability of success \( P(S) = 2x = \frac{2}{3} \).

**Hint:** Set up the relationship between success and failure probabilities based on the given condition.

2. **Use the binomial distribution:**
Let \( n = 6 \) (the number of trials) and \( p = \frac{2}{3} \) (the probability of success). We want to find the probability of at least 4 successes, which is:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) \]
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

3. **Calculate each probability:**
- For \( k = 4 \):
\[ P(X = 4) = \binom{6}{4} \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2 = 15 \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2 \]
- For \( k = 5 \):
\[ P(X = 5) = \binom{6}{5} \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right)^1 = 6 \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right) \]
- For \( k = 6 \):
\[ P(X = 6) = \binom{6}{6} \left( \frac{2}{3} \right)^6 = 1 \left( \frac{2}{3} \right)^6 \]

4. **Combine the probabilities:**
Now, summing these probabilities:
\[ P(X \geq 4) = 15 \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2 + 6 \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right) + \left( \frac{2}{3} \right)^6 \]
Simplifying this gives:
\[ = \frac{15 \cdot 16 + 6 \cdot 8 + 4}{81} \cdot \left( \frac{2}{3} \right)^4 \]
After calculating, we find that this equals \( \frac{31}{9} \left( \frac{2}{3} \right)^{4} \).

5. **Conclusion for Statement I:**
Therefore, Statement I is **True**.

### Statement II:
**The number of times a man must toss a fair coin so that the probability of having at least one head is more than 90% is 4 or more than 4.**

1. **Define the probability:**
Let \( n \) be the number of tosses. The probability of getting at least one head is given by:
\[ P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \left( \frac{1}{2} \right)^n \]

2. **Set up the inequality:**
We want this probability to be greater than 90%:
\[ 1 - \left( \frac{1}{2} \right)^n > 0.9 \]
This simplifies to:
\[ \left( \frac{1}{2} \right)^n < 0.1 \]

3. **Take logarithm:**
Taking logarithm (base 10) on both sides:
\[ n \log_{10} \left( \frac{1}{2} \right) < \log_{10}(0.1) \]
Since \( \log_{10} \left( \frac{1}{2} \right) \) is negative, we can flip the inequality:
\[ n > \frac{\log_{10}(0.1)}{\log_{10} \left( \frac{1}{2} \right)} \]

4. **Calculate the values:**
We know \( \log_{10}(0.1) = -1 \) and \( \log_{10} \left( \frac{1}{2} \right) \approx -0.301 \):
\[ n > \frac{-1}{-0.301} \approx 3.32 \]

5. **Conclusion for Statement II:**
Therefore, \( n \) must be at least 4. Thus, Statement II is also **True**.

### Final Conclusion:
Both statements are correct. The answer is that both statements are true.