The probability a gambler will win money on a certain slot machine is 2.5%. Determine the probability that a gambler will win at least 5 times if they play 100 times.
The correct answer and explanation is :
To solve this problem, we are dealing with a binomial distribution. In a binomial distribution, there are a fixed number of independent trials (in this case, 100 plays), each with two possible outcomes (win or not win), and a constant probability of success (win) in each trial.
Step 1: Defining the parameters of the binomial distribution
We have the following parameters:
- n = 100 (the number of trials, or the number of times the gambler plays the slot machine).
- p = 0.025 (the probability of winning in a single trial, as given by 2.5%).
- We are asked to determine the probability of winning at least 5 times, meaning we want to find $P(X \geq 5)$, where $X$ is the number of wins in 100 plays.
Step 2: Applying the binomial distribution
The number of wins $X$ follows a binomial distribution:
$$
X \sim B(n=100, p=0.025)
$$
The probability mass function (PMF) for a binomial distribution is given by:
$$
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
$$
where $\binom{n}{k}$ is the binomial coefficient, and $k$ is the number of successes (wins).
Step 3: Calculating the probability of winning at least 5 times
The problem asks for $P(X \geq 5)$, which is the probability of having 5 or more wins. We can calculate this by finding the complement, which is the probability of having fewer than 5 wins, and subtracting it from 1:
$$
P(X \geq 5) = 1 – P(X < 5) = 1 – (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4))
$$
This requires calculating the individual probabilities for $P(X = 0), P(X = 1), P(X = 2), P(X = 3),$ and $P(X = 4)$, then summing them up and subtracting from 1.
Step 4: Using a normal approximation
For large $n$, a binomial distribution can be approximated by a normal distribution. Since $n$ is large, we can use the normal approximation to the binomial distribution:
$$
X \sim N(\mu = np, \sigma^2 = np(1-p))
$$
where:
- $\mu = np = 100 \times 0.025 = 2.5$
- $\sigma^2 = np(1-p) = 100 \times 0.025 \times 0.975 = 2.4375$
- $\sigma \approx 1.56$ (the standard deviation)
Using the normal approximation, we convert the discrete probability to a continuous probability using a continuity correction. We want $P(X \geq 5)$, which is equivalent to $P(X \geq 4.5)$ in the normal distribution.
We can standardize this to a z-score:
$$
z = \frac{X – \mu}{\sigma} = \frac{4.5 – 2.5}{1.56} \approx 1.28
$$
Using standard normal tables or a calculator, we find the probability corresponding to a z-score of 1.28, which is approximately 0.8997.
Thus:
$$
P(X \geq 5) \approx 1 – 0.8997 = 0.1003
$$
Step 5: Conclusion
The probability that a gambler will win at least 5 times if they play 100 times is approximately 0.1003, or 10.03%. This means there is about a 10% chance the gambler will win at least 5 times after 100 plays, based on the given probability of winning in each play.