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Q26E

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Found in: Page 467

### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

# Question: Let E be the event that a randomly generated bit string of length three contains an odd number of 1s, and let F be the event that the string starts with 1. Are E and F Independent?

The two events E and F are Independent events as $P\left(E\cap F\right)=P\left(E\right)×P\left(F\right)$ .

See the step by step solution

## Step 1: Given Information

E is the event that a randomly generated bit string of length three contains an odd number of 1s, and F is the event that the string starts with 1

## Step 2: Definition of Independent Events

Independent events are those events whose occurrence is not dependent on any other event.

Two events are Independent if the equation ${\mathbit{P}}\mathbf{\left(}\mathbf{E}\mathbf{\cap }\mathbf{F}\mathbf{\right)}{\mathbf{=}}{\mathbit{P}}\mathbf{\left(}\mathbf{E}\mathbf{\right)}{\mathbf{×}}{\mathbit{P}}\mathbf{\left(}\mathbf{F}\mathbf{\right)}$ holds true.

## Step 1: Given Information

E is the event that a randomly generated bit string of length three contains an odd number of 1s, and F is the event that the string starts with 1.

## Step 3: Calculating the Probability

We know that, E is the event that a randomly generated bit string of length three contains an odd number of 1s.

In general, a bit has a single binary value, either 0 or 1.

Here, totally four bit strings of length three that contains an odd of 1s; 100, 010, 001, 111.

So, the event E becomes,

$E=\left\{100,010,001,111\right\}\phantom{\rule{0ex}{0ex}}and\left|E\right|=4\right]$

Consider F be the event that the string starts with ‘1’.

Here, totally four bit strings of length three that starts with 1; 100, 110, 101, 111.

So, the event F becomes,

$F=\left\{100,110,101,111\right\}\phantom{\rule{0ex}{0ex}}and\left|F\right|=4$

Next we find $E\cap F$ as

$E\cap F=\left\{100,010,001,111\right\}\cap \left\{100,110,101,111\right\}\phantom{\rule{0ex}{0ex}}=\left\{100,111\right\}\phantom{\rule{0ex}{0ex}}and,\left|E\cap F\right|=2$

In general, the sample space S of an experiment is the set of all possible outcomes.

In this experiment the set of all possible outcomes are 8.

Hence, they are, $\left\{000,010,001,100,110,011,101,111\right\}$

So, $\left|S\right|=8$

## Step 4: Determining whether the events E and F are Independent or not

Now, we determine that the events E and F are Independent or not, from the definition of Probability.

Suppose that E is an event that is a subset of sample space S, then the probability of an event E is defined as,

$P\left(E\right)=\frac{\left|E\right|}{\left|S\right|}-------\left(1\right)$ Where, $\left|E\right|$ is the number of outcomes in E, $\left|S\right|$ is the total number of outcomes.

Now, from the definition of Independent sets, Let E and F are two events, then the events E and F are Independent only and only if

$P\left(E\cap S\right)=P\left|E\right|×P\left|S\right|------\left(2\right)$

Using (1) we find $P\left|E\right|$ and $P\left|F\right|$ as,

$P\left(E\right)=\frac{\left|E\right|}{\left|S\right|}=\frac{4}{8}=\frac{1}{2}and\phantom{\rule{0ex}{0ex}}P\left(F\right)=\frac{\left|F\right|}{\left|S\right|}=\frac{4}{8}=\frac{1}{2}$

Next we find

$P\left(E\cap F\right)=\frac{\left|E\cap F\right|}{\left|S\right|}=\frac{2}{8}=\frac{1}{4}$

We can write this as,

$P\left(E\cap F\right)=\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2×2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{1}{2}\phantom{\rule{0ex}{0ex}}=P\left(E\right)×P\left(F\right)$

Hence, $P\left(E\cap F\right)=P\left(E\right)×P\left(F\right)$ .

Therefore, the two events E and F are Independent.