Question

If 4 letters are put at random into 4 envelopes, what is the probability that at least one letter gets into the correct envelope?

Short answer

The limits for $90\%$confidence interval is$1.65\sigma$for$95\%$ is$1.96$and for$99\%$is$2.58$. The interval$\mu \pm 1.3\sigma$has confidence equal

Step-by-step solution

Given Information.

It has been given that 4 letters are put at random into 4 envelopes

Definition of Probability.

Probability means the chances of any event to occur is called it probability.

 Step 3: Find the probability.

Let$X~N(\mu ,{\sigma }^2).$The precent of the confidence interval$\mu \pm h$is equal to the expression mentioned below.

$\begin{array}{c}P(\mu -h\le X\le \mu +h)=P(-\frac{h}{\sigma }\le Z\le \frac{h}{\sigma })\\ =2\Phi \left(\frac{h}{\sigma }\right)-1\end{array}$

Here$Z~N(0,1)$and$\Phi$is the distribution function of Z. Use the programming language (like R) we can calculate the required. For example, the limits$90\%$confidence interval are equal to the expression derived below.

$\begin{array}{c}2\Phi \left(\frac{h}{\sigma }\right)-1=0.9\\ \Phi \left(\frac{h}{\sigma }\right)=0.95\\ h=1.6449\sigma \end{array}$

Here the code is 0.95 similarly, the limits for$95\%$confidence interval is mentioned below.

data-custom-editor="chemistry" $\begin{array}{c}2\Phi \left(\frac{h}{\sigma }\right)-1=0.95\\ \Phi \left(\frac{h}{\sigma }\right)=0.975\\ h=1.96\sigma \end{array}$

The limits$for99\%$confidence interval is given below.

$\begin{array}{c}2\Phi \left(\frac{h}{\sigma }\right)-1=0.99\\ \Phi \left(\frac{h}{\sigma }\right)=0.995\\ h=2.5758\sigma .\end{array}$

Plug $h=1.3\sigma$, the confidence of that interval is equal to the value given below.

$\begin{array}{c}2\Phi \left(\frac{1.3\sigma }{\sigma }\right)-1=2\Phi \left(1.3\right)-1\\ =0.8064\end{array}$