Question

A missile is destroyed before hitting the target by a probability of \(\frac{1}{3}\). If the missile is not destroyed probability of hitting the target is \(\frac{3}{4}\). Three missiles are fired. The probabality of exactly two hits the target is (1) \(\frac{1}{16}\) (2) \(\frac{3}{16}\) (3) \(\frac{3}{8}\) (4) \(\frac{1}{8}\)

Short answer

The probability is \(\frac{3}{8}\). The correct option is (3).

Step-by-step solution

- Calculate the Probability of Hitting the Target

First, determine the probability that a single missile hits the target. The missile is destroyed with a probability of \(\frac{1}{3}\), so the probability that it is not destroyed is \(\frac{2}{3}\). Given it is not destroyed, the probability it hits the target is \(\frac{3}{4}\). Hence, the probability of a missile hitting the target is \[\frac{2}{3} \times \frac{3}{4} = \frac{1}{2}.\]

- Calculate the Probability of Missing the Target

Next, find the probability that a single missile does not hit the target. The probability of hitting the target is \(\frac{1}{2}\), so the probability of missing the target is \[\frac{1}{2}.\]

- Use Binomial Distribution Formula

Since we are dealing with exactly two missiles hitting the target out of three, we use the binomial distribution formula \[\binom{n}{k} p^k (1-p)^{n-k},\] where \( n = 3 \) (number of trials), \( k = 2 \) (number of successes), and \( p = \frac{1}{2} \) (probability of success).

- Calculate Binomial Coefficient

The binomial coefficient (\( \binom{n}{k} \)) is calculated as \[\binom{3}{2} = \frac{3!}{2!(3-2)!} = 3.\] Now substitute this, \( p = \frac{1}{2} \), and \( k = 2 \) into the formula: \[ \binom{3}{2} \times \frac{1}{2}^2 \times \frac{1}{2}^1 = 3 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{8}. \]

Key concepts

Binomial Distribution

The binomial distribution is a fundamental concept in probability theory. It models the number of successes in a fixed number of independent trials, where each trial has the same probability of success. Imagine tossing a fair coin. If you toss it 10 times and want to know the probability of it landing heads exactly 6 times, you'd use the binomial distribution.
The formula for the binomial distribution is: \ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} where:

• n: number of trials or attempts
• k: number of successes
• p: probability of success in each trial
• (1-p): probability of failure in each trial

In our missile exercise, we're interested in finding out the probability of exactly two missiles hitting the target out of three. Hence, we used the binomial distribution formula with n=3, k=2, and p=\(\frac{1}{2}\).

Probability of Independent Events

Independent events are events where the outcome of one event does not affect the outcome of another. This is a significant concept in probability because it simplifies the calculations.
For example, when you calculate the probability of a missile hitting the target, you consider two probabilities:

• the missile not being destroyed, and
• the missile hitting the target if it is not destroyed

These are independent events. The probability of hitting is a product of these individual probabilities: \ \ P(hit) = P(not destroyed) \times P(hit | not destroyed)
In the problem, P(not destroyed) is \(\frac{2}{3}\) and P(hit | not destroyed) is \(\frac{3}{4}\). Thus, the combined probability is \(\frac{2}{3} \times \frac{3}{4} = \frac{1}{2}\).

Combinatorial Calculations

Combinatorial calculations involve determining the number of ways in which events can occur. This is crucial when working with binomial distributions. One common tool is the binomial coefficient, \(\binom{n}{k}\), which calculates the number of ways to choose k successes from n trials.
The formula for the binomial coefficient is: \ \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)
Here, \(!\) denotes a factorial, which is the product of an integer and all the integers below it. For instance, \(3! = 3 \times 2 \times 1 = 6\).

In our exercise, we needed to find \(\binom{3}{2}\), the number of ways to get exactly two hits in three tries. This calculated to: \ \(\binom{3}{2} = \frac{3!}{2!1!} = 3\)
Using this in the binomial formula helped us determine the probability of exactly two hits.