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 :
The problem asks to find the probability that a gambler will win at least 5 times when playing a slot machine 100 times, with the probability of winning on each individual play being 2.5%. This is a binomial probability problem, where we can use the binomial distribution to solve it.
Step 1: Define the binomial distribution parameters
In a binomial distribution, the probability of a specific outcome occurring $x$ times in $n$ trials is given by the formula:
$$
P(X = x) = \binom{n}{x} p^x (1 – p)^{n – x}
$$
Where:
- $n$ is the number of trials (in this case, 100),
- $x$ is the number of successes (wins),
- $p$ is the probability of success on each trial (in this case, 0.025),
- $1 – p$ is the probability of failure on each trial (in this case, 0.975),
- $\binom{n}{x}$ is the binomial coefficient, which represents the number of ways to choose $x$ successes from $n$ trials.
Step 2: Calculate the probability of at least 5 wins
We are interested in the probability of getting at least 5 wins, which means the probability that the number of wins $X$ is 5 or more. This is equivalent to:
$$
P(X \geq 5) = 1 – P(X < 5)
$$
Where $P(X < 5)$ represents the cumulative probability of getting fewer than 5 wins, i.e., $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$.
Step 3: Use a binomial distribution calculator
Given the large number of trials (100), manually calculating the binomial probabilities for each value of $X$ can be tedious. Using a binomial calculator or statistical software (like R or Python), you can compute:
$$
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
$$
Then, subtract this from 1 to find $P(X \geq 5)$.
Step 4: Use a normal approximation (optional)
For large $n$, a normal approximation can be used to simplify the calculation. The binomial distribution can be approximated by a normal distribution with mean $\mu = np$ and standard deviation $\sigma = \sqrt{np(1-p)}$.
For $n = 100$ and $p = 0.025$:
$$
\mu = 100 \times 0.025 = 2.5
$$
$$
\sigma = \sqrt{100 \times 0.025 \times 0.975} \approx 1.537
$$
To approximate $P(X \geq 5)$, we first convert 5 to a z-score:
$$
z = \frac{5 – 2.5}{1.537} \approx 1.62
$$
Using a standard normal table or a calculator, the probability corresponding to $z = 1.62$ is approximately 0.947. Therefore, the probability of having at least 5 wins is:
$$
P(X \geq 5) = 1 – P(Z < 1.62) = 1 – 0.947 = 0.053
$$
Final Answer:
Thus, the probability that the gambler will win at least 5 times is approximately 0.053 or 5.3%.
This result shows that the chances of winning at least 5 times in 100 plays is relatively low, which makes sense given the low probability of winning on any single play (2.5%). The binomial distribution or normal approximation method provides a straightforward way to assess such probabilities for large numbers of trials.