Learning Lab

Example

Suppose that in a particular family the probability that a child will have red hair is \(1/4.\)

Then, if the parents have three children:

1. the probability that all three will have red hair is \[\begin{align*} P(\textrm{Red and Red and Red}) & =P(R_{1}\cap R_{2}\cap R_{3})\\ & =\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}\\ & =\frac{1}{64}. \end{align*}\]

2. the probability that none will have red hair is \[\begin{align*} P(\overline{R}_{1}\cap\overline{R}_{2}\cap\overline{R}_{3}) & =\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\\ & =\frac{27}{64}. \end{align*}\]

3. the probability that at least one child will have red hair is 11 Note: “at least one child has red hair” and “no children have red hair” are complementary events. \[\begin{align*} 1-P(\textrm{none with red hair}) & =1-P(\overline{R}_{1}\cap\overline{R}_{2}\cap\overline{R}_{3})\\ & =1\text{\textendash}\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4}\\ & =1-\frac{27}{64}\\ & =\frac{37}{64}. \end{align*}\]

4. the probability that only the first child will have red hair is \[\begin{align*} P(R_{1}\textrm{ and }\overline{R}_{2}\textrm{ and }\overline{R}_{3}) & =\frac{1}{4}\times\frac{3}{4}\times\frac{3}{4}\\ & =\frac{9}{64}. \end{align*}\]

5. the probability that exactly one child will have red hair is 22 This example demonstrates that it is important to consider all the ways in which the child with red hair might be selected. \[\begin{align*} & P(R_{1}\textrm{ and }\overline{R}_{2}\textrm{ and }\overline{R}_{3})+P(\overline{R}_{1}\textrm{ and }R_{2}\textrm{ and }\overline{R}_{3})+P(\overline{R}_{1}\textrm{ and }\overline{R}_{2}\textrm{ and }R_{3})\\ & =\left(\frac{1}{4}\times\frac{3}{4}\times\frac{3}{4}\right)+\left(\frac{3}{4}\times\frac{1}{4}\times\frac{3}{4}\right)+\left(\frac{3}{4}\times\frac{3}{4}\times\frac{1}{4}\right)\\ & =3\times\left(\frac{1}{4}\right)^{1}\left(\frac{3}{4}\right)^{2}\\ & =3\times\frac{9}{64}\\ & =\frac{27}{64}. \end{align*}\]

Example

One in every hundred items a machine produces are defective. What is the probability that in a sample of five items produced by this machine
(a) Exactly three are defective?
(b) None are defective ?
(c) At least 1 is defective?

Solution:

We have

\(n=5,\quad p=\frac{1}{100}=0.01,\quad x=3\)

1. Using equation \(\left(1\right)\) from above \[\begin{align*} P(X=x) & =\textrm{}^{n}C\textrm{}_{x}\times p^{x}\times\textrm{$(1-p)^{n-x}$ }\\ P(X=3) & =\textrm{}^{5}C\textrm{}_{3}\times0.01^{3}\times\textrm{$(1-0.01)^{5-3}$ }\\ & \approx0.00001. \end{align*}\] So if we obtained three defective items in a sample of \(5\) we might be suspicious of the claim that only one in a hundred is defective!

2. Using equation \(\left(1\right)\) from above \[\begin{align*} P(X=0) & =\textrm{}^{5}C\textrm{}_{0}\times0.01^{0}\times\textrm{$(1-0.01)^{5}$ }\\ & \approx0.95\,. \end{align*}\]

3. Remember “none defective” and “at least one defective” are complementary events. So

\[\begin{align*} P(X\geq1) & =1-P(X=0)\\ & =1-0.95\\ & =0.05. \end{align*}\]