The text provided in the image is in Gujarati. Here's the translation to English:

"In a game, a coin is tossed 3 times, and the results of each toss are noted. If the results of all three tosses are heads, or all three are tails, the player wins. Otherwise, the player loses. Calculate the probability of the player winning."

This problem is related to the concept of probability and involves multiple coin tosses.

Solution:

When a coin is tossed 3 times, the total number of possible outcomes is $2^3 = 8$, because each toss has 2 possible outcomes (heads or tails).

The player wins if all tosses result in either heads (HHH) or tails (TTT). There are only two such outcomes:

1. HHH (All heads)
2. TTT (All tails)

Thus, the number of favorable outcomes for winning is 2.

Therefore, the probability of winning is the ratio of favorable outcomes to the total outcomes:

$\text{Probability of winning} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{8} = \frac{1}{4}$

Final Answer:

The probability of winning the game is $\frac{1}{4}$ or 0.25.

Do you have any questions or need further details?

Here are some related questions:

1. What is the probability of losing the game?
2. How would the probability change if the coin is tossed 4 times?
3. Can this problem be modeled using a binomial distribution?
4. What if the player wins when there are exactly 2 heads out of 3 tosses?
5. How would the probability change if the coin were biased?

Tip: For any probability problem, always count the total number of possible outcomes first, and then identify the favorable outcomes.