Question

(a) Three typed letters and their envelopes are piled on a desk. If someone puts theletters into the envelopes at random (one letter in each), what is theprobabilitythat each letter gets into its own envelope? Call the envelopes A, B, C, and thecorresponding letters a, b, c, and set up the sample space. Note that “a in A,b in B, c in A” is one point in the sample space.

(b) What is the probability that at least one letter gets into its own envelope?

Hint: What is the probability that no letter gets into its own envelope?

(c) Let A mean that a got into envelope A, and so on. Find the probability P(A)that a got into A. Find P(B) and P(C). Find the probability P(A + B)that either a or b or both got into their correct envelopes, and the probabilityP(AB) that both got into their correct envelopes. Verify equation (3.6).

Short answer

Answer

(a) The sample space is $$\begin{gathered} ainAbinBcinC \\ ainAbinCcinB \\ ainCbinBcinA \\ ainBbinAcinC \\ ainBbinCcinA \\ ainCbinAcinB \end{gathered}$$and the probability that each letter gets into its own envelope is$\frac{1}{6}$.

(b) The probability that at least one letter gets into its own envelope is $\frac{2}{3}$.

(c) $P\left(A\right)=P\left(B\right)=P\left(C\right)=\frac{1}{3},P\left(A+B\right)-\frac{1}{3},P\left(A+B\right)-\frac{1}{2},P\left(AB\right)=\frac{1}{6}$and equation 3.6 is valid.

Step-by-step solution

Given Information

A , B and C are the envelopes and a , b and Cc are the corresponding letters ,the letters are to be put inside the envelopes.

Definition of Independent Event

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

When events are independent, apply the formula $P\left(AB\right)=P\left(A\right)·P\left(B\right)$where A and B are the events.

Creating The sample space

Create the sample space for the given experiment.

$$\begin{gathered} ainAbinBcinC \\ ainAbinCcinB \\ ainCbinBcinA \\ ainBbinAcinC \\ ainBbinCcinA \\ ainCbinAcinB \end{gathered}$$

Finding the probability that each letter gets into its own envelope

Each point of the sample space has the probability $\frac{1}{6}$and only one point satisfies the condition that the correct letter pair and thus has probability$\frac{1}{6}$.

Finding the probability that at least one letter gets into its own envelope

To find at least one letter gets into its own envelope we can find, that no one gets into the correct envelope and subtract the probability from 1 to get the required probability.

It can be observed that there are only two point in which all the letter envelop pairs are wrong and thus the probability is $\frac{1}{3}$.

This implies that the probability that at least one letter gets into its own envelope is $1-\frac{1}{3}=\frac{2}{3}$.

Finding the probabilityP(A)  ,  P(B) and P(C) Find the probability P(A+B) that either a or b or both got into their correct envelopes, and the probability P(AB) that both got into their correct envelopes

Let A mean that letter “a” got into envelope “A” ,B mean that letter “b” got into envelope “B” and mean that letter “c” got into envelope “C”.

It can be observed that only 2 point satisfies each event A, B and C, this implies that$P\left(A\right)=P\left(B\right)=P\left(C\right)=\frac{1}{3}$.

It can be observed that there are only 3 point where letter “a” got into envelope “A” or letter “b” got into envelope “B” or both these implies that$P\left(AB\right)=\frac{1}{6}$.

It can be observed that there is only 1 point where letter “a” got into envelope “A” and letter “b” got into envelope “B” this implies that $P\left(AB\right)=\frac{1}{6}$.

Substitute the obtained values $P\left(A\right),P\left(B\right)$and $P\left(AB\right)$of into the right side of formula $P\left(A+B\right)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$.

$$\begin{gathered} \frac{1}{3}+\frac{1}{3}-\frac{1}{6}=\frac{2+2-1}{6} \\ =\frac{1}{2} \end{gathered}$$

This show that equation 3.6 is valid.