Understanding Probability of At Least One Hit

The problem asks for the probability of getting at least one hit when a person takes 4 shots at a target, given that the probability of hitting the target in a single shot is 0.5.

Probability of a Single Shot Outcome

Let H denote the event of hitting the target in a single shot, and M denote the event of missing the target in a single shot.

• The probability of hitting the target is given as \(P(H) = 0.5\).
• The probability of missing the target is \(P(M) = 1 - P(H)\).
• So, \(P(M) = 1 - 0.5 = 0.5\).

Each shot is assumed to be independent of the others.

Calculating Probability of At Least One Hit

The event "at least one hit in 4 shots" includes the possibilities of 1 hit, 2 hits, 3 hits, or 4 hits. Calculating the probability for each of these cases and summing them up can be complex.

A simpler approach is to use the concept of complementary probability. The complementary event to "at least one hit in 4 shots" is "no hits in 4 shots".

The probability of "at least one hit" is equal to 1 minus the probability of "no hits".

Probability (at least one hit) = 1 - Probability (no hits in 4 shots)

Probability of No Hits in 4 Shots

Getting "no hits in 4 shots" means the person misses the target on all 4 shots. Since the shots are independent, the probability of this sequence of events is the product of the probabilities of missing each individual shot:

Probability (no hits in 4 shots) = \(P(M \text{ on shot 1}) \times P(M \text{ on shot 2}) \times P(M \text{ on shot 3}) \times P(M \text{ on shot 4})\)

Since \(P(M) = 0.5\) for each shot:

Probability (no hits in 4 shots) = \(0.5 \times 0.5 \times 0.5 \times 0.5\)

This can be written as \((0.5)^4\).

Let's calculate the value:

\((0.5)^4 = \left(\frac{1}{2}\right)^4 = \frac{1^4}{2^4} = \frac{1}{16}\)

Final Probability Calculation

Now, we can find the probability of at least one hit:

Probability (at least one hit) = 1 - Probability (no hits in 4 shots)

Probability (at least one hit) = \(1 - \frac{1}{16}\)

To subtract, we find a common denominator:

\(1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{16 - 1}{16} = \frac{15}{16}\)

So, the probability of getting at least one hit in 4 shots is \(\frac{15}{16}\).

Comparing with Options

Let's compare our calculated probability with the given options:

• Option 1: \(\frac{1}{8}\)
• Option 2: \(\frac{1}{16}\)
• Option 3: \(\frac{15}{16}\)
• Option 4: \(\frac{7}{8}\)

Our result, \(\frac{15}{16}\), matches Option 3.

Revision Table: Key Probability Concepts

Additional Information: Bernoulli Trials

This problem involves a sequence of independent trials (the 4 shots), where each trial has only two possible outcomes (hit or miss), and the probability of success (hit) is constant for each trial. This is an example of a Bernoulli trial sequence.

For a fixed number of Bernoulli trials (n=4 in this case) with a constant probability of success (p=0.5), the number of successes (hits) follows a Binomial distribution. The probability of getting exactly k successes in n trials is given by:

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

Where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

While we solved the "at least one hit" problem using the complementary event (which is simpler here), we could also solve it by summing the probabilities of 1, 2, 3, or 4 hits:

\(P(\text{at least one hit}) = P(X=1) + P(X=2) + P(X=3) + P(X=4)\)

Let's quickly check one term, for instance, the probability of exactly 1 hit in 4 shots:

\(P(X=1) = \binom{4}{1} (0.5)^1 (0.5)^{4-1} = 4 \times 0.5 \times (0.5)^3 = 4 \times 0.5 \times 0.125 = 4 \times 0.0625 = 0.25 = \frac{1}{4}\)

Calculating and summing all terms \(P(X=1)\) through \(P(X=4)\) would also give \(\frac{15}{16}\), but the complementary probability approach is more efficient.