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Suppose the insurance agency from chapter 4.2 is conducting a random sample of four individuals. What is the chance that exactly one of them will exceed the deductible?

$$ \begin{align} P (A&=\text{ exceed, }B=\text{ not, }C= \text{ not, }D=\text{not}) \ &=P(A=\text{ exceed})P(B=\text{ not})P(C=\text{not})P(D=\text{not}) \ &= (0.3)(0.7)(0.7)(0.7)\ &= (0.7)3(0.3)1\ &= 0.103

\end{align} $$

But there are three other scenarios: B, C, or D could exceed while the rest don’t. So the total probability is actually \(4 \times (0.7)^3(0.3)^1=0.412\)

Scenarios like this are examples of binomial distribution.

The general formula for the number of ways to choose \(k\) successes in \(n\) trials (i.e. arrange \(k\) successes and \(n-k\) failures) is

\[ \large \binom{n}{k}=\frac{n!}{k!(n-k)!} \]

\(\binom{n}{k}\) is read as \(n\) choose \(k\).

Example:

\[ \binom{4}{3}=\frac{4!}{3!(4-3)!}=\frac{4!}{3!1!}=\frac{(4)(3)(2)(1)}{((3)(2)(1))(1)}=\frac{28}{6}=4 \]

If the example allows repetition, we use this formula instead:

\[ C(n+k-1, k)=\frac{(n+k-1)!}{k!(n-1)!} \]

💡 Binomial Distribution Suppose the probability of a single trial being a success is \(p\). Then the probability of observing exactly \(k\) successes in \(n\) independent trials is given by \(\large \binom{n}{k}p^k(1-p)^{n-k}=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\) The mean is \(\mu = np\) The variance is \(\sigma^2=np(1-p)\), and the standard deviation is the square root of this.

Is it binomial?

  • Are trials independent?
  • Is the number of trials (n) fixed?
  • Can each trial outcome be classified as a success or failure?
  • Is the probability of success (p) the same for each trial?

Normal Approximation to the Binomial Distribution

The binomial formula is cumbersome when the sample size is large. In some cases we can use the normal distribution as an easier and faster way to estimate binomial probabilities.

💡 Normal Approximation of the Binomial Distribution The binomial distribution with a probability of success \(p\) is nearly normal when the sample size \(n\) is sufficiently large that \(np\) and \(n(1-p)\) are both at least 10. The approximate normal distribution has parameters corresponding to the mean and standard deviation of the binomial distribution: \(\mu=np\) \(\sigma=\sqrt{np(1-p)}\)