To find the probability of winning exactly 7 games out of 50 in a ring toss game where the probability of winning a single game is 10% (or 0.10), we can approximate the binomial distribution using the normal distribution. This method works because the number of trials (50) is large.

Step 1: Parameters of the Binomial Distribution

The binomial distribution is defined by two parameters:

• $n = 50$ (the number of games played),
• $p = 0.10$ (the probability of winning a single game).

Step 2: Mean and Standard Deviation of the Binomial Distribution

For a binomial distribution, the mean ($\mu$) and the standard deviation ($\sigma$) are given by:

• Mean: $\mu = np = 50 \times 0.10 = 5$,
• Standard deviation: $\sigma = \sqrt{np(1 - p)} = \sqrt{50 \times 0.10 \times 0.90} = \sqrt{4.5} \approx 2.121$.

Step 3: Normal Approximation

We approximate the binomial distribution $B(n, p)$ with a normal distribution $N(\mu, \sigma^2)$. Therefore, we approximate the binomial distribution with the normal distribution $N(5, 2.121^2)$.

Step 4: Apply the Continuity Correction

Since the binomial distribution is discrete, but the normal distribution is continuous, we use a continuity correction. To find the probability of winning exactly 7 games, we calculate the probability of winning between 6.5 and 7.5 games (i.e., the interval $6.5 \leq X \leq 7.5$).

Step 5: Standardize the Values (Convert to Z-scores)

The Z-score formula is: $Z = \frac{X - \mu}{\sigma}$ We need to calculate the Z-scores for $X = 6.5$ and $X = 7.5$.

For $X = 6.5$: $Z_1 = \frac{6.5 - 5}{2.121} \approx \frac{1.5}{2.121} \approx 0.707$

For $X = 7.5$: $Z_2 = \frac{7.5 - 5}{2.121} \approx \frac{2.5}{2.121} \approx 1.179$

Step 6: Use the Standard Normal Distribution

Now, we need to look up the Z-scores in the standard normal table or use a calculator to find the corresponding probabilities.

For $Z_1 = 0.707$, the cumulative probability is approximately 0.76. For $Z_2 = 1.179$, the cumulative probability is approximately 0.881.

Step 7: Calculate the Probability

The probability of winning between 6.5 and 7.5 games is: $P(6.5 \leq X \leq 7.5) = P(Z_2) - P(Z_1) = 0.881 - 0.76 = 0.121$

Final Answer:

The probability of winning exactly 7 games out of 50 is approximately 0.121, or 12.1%.

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Would you like more details or explanations on any step?

Here are 5 related questions:

1. How would you calculate the probability of winning more than 7 games out of 50?
2. How does the continuity correction improve the approximation of the binomial distribution by the normal distribution?
3. What is the binomial distribution formula and how is it used for small sample sizes?
4. How would the result change if the probability of winning per game was different, say 20%?
5. How do we apply the normal approximation for extreme values (e.g., when $p$ is very close to 0 or 1)?

Tip: The normal approximation works well for binomial distributions when both $np$ and $n(1 - p)$ are greater than 5, which ensures the distribution is not too skewed.