Testing a watch manufacturer’s claim. A manufacturer of a new Smart Watch claims that the probability of its watch running more than 1 minute slow or 1 minute fast after 1 year of use is .05. A consumer protection agency has purchased four of the manufacturer’s watches with the intention of testing the claim.

1. Assuming that the manufacturer’s claim is correct, what is the probability that none of the watches are as accurate as claimed?
2. Assuming that the manufacturer’s claim is correct, what is the probability that exactly two of the four watches are as accurate as claimed?
3. Suppose that only one of the four tested watches is as accurate as claimed. What inference can be made about the manufacturer’s claim? Explain.
4. Suppose that none of the watches tested are as accurate as claimed. Is it necessarily true that the manufacturer’s claim is false? Explain.

The formula for probability is

$\begin{array}{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{P}\mathbf{=}\frac{\mathrm{Favourable}\mathrm{out}\mathrm{comes}}{\mathrm{Total}\mathrm{Out}\mathrm{comes}}\\ \mathbf{P}\mathbf{(}{\mathbf{A}}^C\mathbf{)}\mathbf{=}\mathbf{1}-\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\end{array}$

Here P(A)=0.05

$\begin{array}{c}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{1}-\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}-\mathbf{0}\mathbf{.}\mathbf{05}\\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{95}\end{array}$

The sample spaces are:

(A, A, A, A), (N,N,N,N) ,(N,A,N,A) ,(A,N,A,N) ,(N,A,A,N), (A,N,A,A) ,(A,A,N,A) ,(A,A,A,N), (N,N,A,A), (A,A,N,N), (N,A,A,N), (A,N,N,A),(N,N,,N,A), (A,N,N,N) ,(N,A,N,N), (N,N,A,N).

MOW,

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\\ \mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00000625}\end{array}$

So, the probability is 0.00000625.

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{exectly}\mathbf{}\mathbf{two}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{N}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{A}\mathbf{)}\mathbf{+}(\mathbf{A}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{N})\mathbf{+}(\mathbf{N}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{N})\mathbf{+}(\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{A})\mathbf{+}(\mathbf{A}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{N})\mathbf{+}(\mathbf{A}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{A})\\ =6\times P(N,A,N,A)\\ \mathbf{=}\mathbf{6}\mathbf{\times }\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\\ \mathbf{=}\mathbf{6}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0135}\end{array}$

Accordingly, the probability is 0.0135.

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{1}\mathbf{watch}\mathbf{}\mathbf{accurate}\mathbf{)}\mathbf{=}(\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{A})\mathbf{+}(\mathbf{A}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{N})\mathbf{+}(\mathbf{N}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{N})\mathbf{+}(\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{,}\mathbf{A}\mathbf{,}\mathbf{N})\\ =4\times P(N,N,N,A)\\ \mathbf{=}\mathbf{4}\mathbf{\times }\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{N}\mathbf{\right)}\mathbf{.}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{95}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{000475}\end{array}$

The conclusion is that the probability is 0.000475, which is so low. as a result, there is skepticism, as the manufacturer said.

The probability of no accurate watches is 0.00000625, which is so low. Therefore, the manufacturers’ claim is not wrong.