Two coins are tossed simultaneously 300 times and it is found that two heads appeared 135 times, one head appeared 111 times and no head appeared 54 times. If two coins are tossed at random, what is the probability of getting (i) 2 heads (ii) 1 head (iii) 0 head?

Answer

To solve the problem, we need to find the probability of getting 2 heads, 1 head, and 0 heads when two coins are tossed. We will use the formula for probability:

\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \]

### Step 1: Calculate the probability of getting 2 heads

Given:
- Number of times 2 heads appeared = 135
- Total number of tosses = 300

Using the probability formula:

\[ P(\text{2 heads}) = \frac{135}{300} \]

Now, we simplify the fraction:

\[ P(\text{2 heads}) = \frac{135 \div 15}{300 \div 15} = \frac{9}{20} \]

### Step 2: Calculate the probability of getting 1 head

Given:
- Number of times 1 head appeared = 111
- Total number of tosses = 300

Using the probability formula:

\[ P(\text{1 head}) = \frac{111}{300} \]

This fraction cannot be simplified further, so:

\[ P(\text{1 head}) = \frac{111}{300} \]

### Step 3: Calculate the probability of getting 0 heads

Given:
- Number of times 0 heads appeared = 54
- Total number of tosses = 300

Using the probability formula:

\[ P(\text{0 heads}) = \frac{54}{300} \]

Now, we simplify the fraction:

\[ P(\text{0 heads}) = \frac{54 \div 6}{300 \div 6} = \frac{9}{50} \]

### Final Answers:

- Probability of getting 2 heads: \( \frac{9}{20} \)
- Probability of getting 1 head: \( \frac{111}{300} \)
- Probability of getting 0 heads: \( \frac{9}{50} \)