Question

Question: Recall from Definition \(5\) in Section \(7.2\) that the events \({E_1}\;,\;{E_2}\;, \ldots ,\;{E_n}\) are mutually independent if \(p\left( {{E_{{i_1}}} \cap {E_{{i_2}}} \cap \cdots \cap {E_{{i_{\rm{m}}}}}} \right) = p\left( {{E_{{i_1}}}} \right)p\left( {{E_{{i_2}}}} \right) \cdots p\left( {{E_{{i_{\rm{m}}}}}} \right)\) whenever \({i_j}\;,\;j = 1\;,\;2\;, \ldots ,\;m\), are integers with \(1 \le {i_1} < {i_2} < \cdots < {i_m} \le n\) and \(m \ge 2\).

a) Write out the conditions required for three events \({E_1}\;,\;{E_2}\), and \({E_3}\) to be mutually independent.

b) Let \({E_1}\;,\;{E_2}\), and \({E_3}\) be the events that the first flip comes up heads, that the second flip comes up tails, and that the third flip comes up tails, respectively, when a fair coin is flipped three times. Are \({E_1}\;,\;{E_2}\), and \({E_3}\) mutually independent?

c) Let \({E_1}\;,\;{E_2}\), and \({E_3}\) be the events that the first flip comes up heads, that the third flip comes up heads, and that an even number of heads come up, respectively, when a fair coin is flipped three times. Are \({E_1}\), \({E_1}\;,\;{E_2}\), and \({E_3}\) pairwise independent? Are they mutually independent?

d) Let \({E_1}\;,\;{E_2}\), and \({E_3}\) be the events that the first flip comes up heads, that the third flip comes up heads, and that exactly one of the first flip and third flip come up heads, respectively, when a fair coin is flipped three times. Are \({E_1}\;,\;{E_2}\), and \({E_3}\) pairwise independent? Are they mutually independent?

e) How many conditions must be checked to show that \(n\) events are mutually independent?

Short answer

Answer

a) The conditions are as follows:

\(\begin{aligned}{}P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_2}} \right)P\left( {{E_3}} \right)\\P\left( {{E_1} \cap {E_2}} \right) &= P\left( {{E_1}} \right)P\left( {{E_2}} \right)\\P\left( {{E_1} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_3}} \right)\\P\left( {{E_2} \cap {E_3}} \right) &= P\left( {{E_2}} \right)P\left( {{E_3}} \right)\end{aligned}\)

b) Yes, \({E_1},{E_2}\) and \({E_3}\) mutually independent as well as pairwise independent.

c) Yes, \({E_1},{E_2}\) and \({E_3}\) are mutually independent as well as pairwise independent.

d) No \({E_1},{E_2}\) and \({E_3}\) are not mutually independent while they are pairwise independent.

e) Number conditions are \({2^n} - n - 1\).

Step-by-step solution

Given data

For first flip, \({E_1}\) be the event that comes up heads, \({E_2}\) be the event that the third flip comes up heads and \({E_3}\) be the event that the even number of heads comes up.

Concept of Probability

If\(S\)is a finite non-empty sample space of equally likely outcomes and\(E\)is an event which is a subset of\(S\).

The probability of\(E\)is\(P(E) = \frac{{\left| E \right|}}{{\left| S \right|}}\).

Calculation for conditions required for three events to be mutually independent

a)

By definition a set of \(n\) events \({E_1},{E_2}, \ldots \ldots .{E_n}\) are independent if \(P\left( {{E_1} \cap {E_2} \cap \ldots \ldots .{E_n}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}} \right) \ldots \ldots ..P\left( {{E_n}} \right)\) and each pair of events are independent.

Thus, if 3 events \({E_1},{E_2}\) and \({E_3}\) to be mutually independent.

Then, the conditions are given as:

\(\begin{aligned}{}P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_2}} \right)P\left( {{E_3}} \right)\\P\left( {{E_1} \cap {E_2}} \right) &= P\left( {{E_1}} \right)P\left( {{E_2}} \right)\\P\left( {{E_1} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_3}} \right)\\P\left( {{E_2} \cap {E_3}} \right) &= P\left( {{E_2}} \right)P\left( {{E_3}} \right)\end{aligned}\)

Calculation to check the three events to be mutually independent

b)

Here, \({E_1} \cap {E_2} \cap {E_3}\) denotes the event that the first flip is a head, the second and third flip are tails. Then, \(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = \frac{1}{8}\).

\(P\left( {{E_1}} \right) = \frac{1}{2}\;,\;P\left( {{E_2}} \right) = \frac{1}{2}\;,\;P\left( {{E_3}} \right) = \frac{1}{2}\)

So, \(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}} \right)P\left( {{E_3}} \right)\).

Here, \({E_1} \cap {E_2}\) denotes the event that the first flip is a head and the second flip is a tail.

Then, \(P\left( {{E_1} \cap {E_2}} \right) = \frac{1}{4} = \) \(P\left( {{E_1}} \right)P\left( {{E_2}} \right)\)

Here, \({E_1} \cap {E_3}\) denotes the event that the first flip is a head and the third flip is a tail.

Then,

\(\begin{aligned}{}P\left( {{E_1} \cap {E_3}} \right) &= \frac{1}{4}\\P\left( {{E_1} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_3}} \right)\end{aligned}\)

Here, \({E_2} \cap {E_3}\) denotes the event that the second flip is a tail and the third flip is a tail. Then

\(\begin{aligned}{}P\left( {{E_2} \cap {E_3}} \right) &= \frac{1}{4}\\P\left( {{E_2} \cap {E_3}} \right) &= P\left( {{E_2}} \right)P\left( {{E_3}} \right)\end{aligned}\)

Hence, the \(3\) events are mutually independent and pairwise mutually independent.

Calculation to check the three events mutually independent as well as pairwise independent

c)

Here, \({E_1} \cap {E_2} \cap {E_3}\) denotes the event that the first flip is a head, the third flip is a head and that we get an even number of heads. This forces that the second flip is a tail and only one possible outcome HTH.

\(\begin{aligned}{}P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) &= \frac{1}{8}\\P\left( {{E_1}} \right) = \frac{1}{2}\;,\;P\left( {{E_2}} \right) &= \frac{1}{2}\;,\;P\left( {{E_3}} \right) &= \frac{1}{2}\end{aligned}\)

So, \(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) = P\left( {{E_1}} \right)P\left( {{E_2}} \right)P\left( {{E_3}} \right)\)

Here, \({E_1} \cap {E_2}\) denotes the event that the first flip is a head and the third flip is a tail.

Then,\(P\left( {{E_1} \cap {E_2}} \right) = \frac{1}{4} = P\left( {{E_1}} \right)P\left( {{E_2}} \right)\).

Here, \({E_1} \cap {E_3}\) denotes the event that the first flip is a head and we get an even number of heads. This leads to only two outcomes HHT and HTH. So, we get:

\(\begin{aligned}{}P\left( {{E_1} \cap {E_3}} \right) &= \frac{1}{4}\\P\left( {{E_1} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_3}} \right)\end{aligned}\)

Here, \({E_2} \cap {E_3}\) denotes the event that the third flip is a head and we get an even number of heads. This leads to only two possible outcomes HTH and THH.

\(\begin{aligned}{}P\left( {{E_2} \cap {E_3}} \right) &= \frac{1}{4}\\P\left( {{E_2} \cap {E_3}} \right) &= P\left( {{E_2}} \right)P\left( {{E_3}} \right)\end{aligned}\)

Hence the \(3\) events are mutually independent and hence pairwise independent.

Calculation to check the three events pairwise independent not mutually

d)

Here, \({E_1} \cap {E_2} \cap {E_3}\) denotes the event that the first flip is a head, the third flip is a head and that exactly one of first and third flip is a head. Clearly this can't be possible hence:

\(\begin{aligned}{}P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) &= 0\\P\left( {{E_1}} \right) = \frac{1}{2}\,,\;P\left( {{E_2}} \right) &= \frac{1}{2}\;,\;P\left( {{E_3}} \right) &= \frac{1}{2}\end{aligned}\)

So \(P\left( {{E_1} \cap {E_2} \cap {E_3}} \right) \ne P\left( {{E_1}} \right)P\left( {{E_2}} \right)P\left( {{E_3}} \right)\)

So, the \(3\) events are not mutually independent.

Here, \({E_1} \cap {E_2}\) denotes the event that the first flip is a head and the third flip is a head.

Then, the condition is given as:

\(\begin{aligned}{}P\left( {{E_1} \cap {E_2}} \right) &= \frac{1}{4}\\P\left( {{E_1} \cap {E_2}} \right) &= P\left( {{E_1}} \right)P\left( {{E_2}} \right)\end{aligned}\)

Here, \({E_1} \cap {E_3}\) denotes the event that the first flip is a head and that exactly one of first and third flip is a heads. This leads to only two outcomes HHT and HTT. So we get:

\(\begin{aligned}{}P\left( {{E_1} \cap {E_3}} \right) &= \frac{1}{4}\\P\left( {{E_1} \cap {E_3}} \right) &= P\left( {{E_1}} \right)P\left( {{E_3}} \right)\end{aligned}\)

Here, \({E_2} \cap {E_3}\) denotes the event that the third flip is a head and that exactly one of first and third flip is a heads. This leads to only two possible outcomes TTH and THH.

\(\begin{aligned}{}P\left( {{E_2} \cap {E_3}} \right) &= \frac{1}{4}\\P\left( {{E_2} \cap {E_3}} \right) &= P\left( {{E_2}} \right)P\left( {{E_3}} \right)\end{aligned}\)

Hence the \(3\) events are pairwise independent.

Calculation to check the conditions for events mutually independent

e)

We have to verify that the \(n\) events are independent two at a time, three at a time and so on up to \(n\) at a time. The number of conditions is the number of ways of choosing two numbers + the number of ways of choosing \(3\) members so on until number of ways of choosing \(n\) members. This is given by Number of conditions to be checked as \(n{C_2} + n{C_3} \ldots \ldots .. + n{C_n}\).

Using that \(n{C_0} + n{C_1} + n{C_2} + n{C_3} \ldots \ldots .. + n{C_n} = {2^n}\).

We get the conditions as:

\(n{C_2} + n{C_3} \ldots \ldots + n{C_n} = {2^n} - n - 1\)