In a game where a player throws a dart at a circular target, assuming every throw hits the target and the probability of hitting any specific region is uniform, what is the probability of hitting a particular region that covers one-fifth of the target area in two out of three throws?

Official Answer

Certainly, engaging with probability questions offers a fascinating insight into analytical thinking and problem-solving skills, which are crucial in my role as a Data Scientist. In tackling the question at hand, we're examining a scenario where a player throws a dart at a circular target three times, with the aim of hitting a specific region that constitutes one-fifth of the total area in exactly two out of those three attempts. Given my background in data analysis and probability theory, let's dissect this problem step by step to uncover the solution, employing a binomial probability formula as our guiding framework.

The essence of this problem lies in understanding the binomial probability distribution, which in simple terms, helps us find the probability of having exactly k successes (hits in our case) in n trials (throws), given the probability of success on a single trial. Here, the probability of hitting the desired region in a single throw is 1/5, as the region covers one-fifth of the target. Conversely, the probability of not hitting the target is 4/5.

To calculate the probability of hitting the specific region exactly twice in three throws, we apply the formula for binomial probability:

[P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}]

where - (P(X = k)) is the probability of having exactly k successes in n trials, - (\binom{n}{k}) is the binomial coefficient, representing the number of combinations for k successes out of n trials, - (p) is the probability of success on a single trial, and - (1-p) is the probability of failure on a single trial.

Substituting the values corresponding to our scenario: - (n = 3) (since the player throws the dart three times), - (k = 2) (since we're interested in the event where the player hits the target region exactly twice), - (p = 1/5) (the probability of hitting the specific region), - (1-p = 4/5) (the probability of not hitting the specific region),

we get:

[P(X = 2) = \binom{3}{2} \cdot \left(\frac{1}{5}\right)^2 \cdot \left(\frac{4}{5}\right)^{3-2}]

Calculating the binomial coefficient (\binom{3}{2} = 3), the formula simplifies to:

[P(X = 2) = 3 \cdot \left(\frac{1}{5}\right)^2 \cdot \left(\frac{4}{5}\right)^1]

[P(X = 2) = 3 \cdot \frac{1}{25} \cdot \frac{4}{5}]

[P(X = 2) = \frac{3 \cdot 4}{125}]

[P(X = 2) = \frac{12}{125}]

Therefore, the probability of hitting a particular region that covers one-fifth of the target area in two out of three throws is (\frac{12}{125}), or 0.096.

In my analytical work as a Data Scientist, breaking down complex problems into solvable components, as demonstrated, has been instrumental in delivering insights and solutions across various projects. This approach not only ensures accuracy but also fosters a deep understanding of underlying principles, enabling the crafting of innovative strategies and algorithms tailored to specific challenges. The ability to articulate this process clearly is vital, empowering stakeholders to appreciate the value of analytical endeavors and make informed decisions.