Question

In Exercises \(\mathrm{P} .108\) to \(\mathrm{P} .111,\) calculate the requested binomial probability. Find \(P(X=3)\) if \(X\) is a binomial random variable with \(n=10\) and \(p=0.4\)

Short answer

The probability that the binomial random variable \(X\) takes the value 3 is approximately 0.2149.

Step-by-step solution

Identify the relevant parameters

Here, the number of trials \(n\) is 10, the value of the random variable \(X\) we're interested in \(k\) is 3, and the probability of success on each trial \(p\) is 0.4.

Substitute values into the binomial formula

We substitute these values into the binomial distribution formula to get \[P(X=3)= \binom{10}{3} \times (0.4)^3 \times (1-0.4)^{10-3}\]

Evaluate the binomial coefficient

Next, we evaluate the binomial coefficient, which is the number of ways of choosing 3 successes from 10 trials. In mathematics, this is written as \(\binom{10}{3}\), and equals \[ \frac{10!}{3! \times (10-3)!}\] where '!' denotes the factorial function. Simplifying this, we get 120.

Evaluate the remaining expression

Lastly, we have \[120 \times (0.4)^3 \times (1-0.4)^7\], which if rounded to four decimal places equals 0.2149.