# Conditional probability of event $E$ given $F$ [closed]

Let $E$ be the event that a randomly generated bit string of length $5$ starts with $0$, and let $F$ be the event that a randomly generated bit string of length $5$ has an odd number of $1’s$.

What is the conditional probability that a randomly generated bit string of length $5$ starts with $0$ given that the string has an odd number of $1’s$, i.e. the conditional probability of event $E$ given $F$? Are $E$ and $F$ independent events?

## closed as off-topic by Did, Mostafa Ayaz, Vladhagen, Delta-u, José Carlos SantosSep 24 '18 at 17:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Mostafa Ayaz, Vladhagen, Delta-u, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

Assume that all the outcomes are equally probable; let the probability of each outcomes be $\frac 1{32}$.

Then $$\#E=\#\{00001,00010,00011,00100,00101,00110,00111,...,01111\}=16.$$

and

$$\#F=\#\{00001,00010,00100,00111,01000,01011,...,11111\}=16$$

and

$$\#E\cap F=\#\{00001,00010,00100,00111,01000,01011,01101,01110\}=8.$$

The conditional probability in question is

$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=$$ $$=\frac{P(\{00001,00010,00100,00111,01000,01011,01101,01110\})}{P(\{00001,00010,00100,00111,10000,01011,11001,11100,...,11111\})}=$$ $$=\frac{8/32}{16/32}=\frac12.$$

$E$ and $F$ are independent because $P(E\cap F)=\frac8{32}=\frac14$, $P(E)=\frac12$ and $P(F)=\frac{16}{32}=\frac12$ and $\frac12\times \frac12=\frac 14$

• @paw88789: Do you mean $P(E\mid F)=\frac 1{32}$? – zoli Sep 27 '15 at 11:40
• The outcomes for $F \cap E$ are $01000, 00100,00010,00001,01110,01101,01011,00111$. Therefore $P(F \cap E)=\frac{8}{32}=\frac{1}{4}$ From where does 64 come from, since $2^5=32$ ? – callculus Sep 27 '15 at 11:55
• @calculus: Sure, I have to edit quickly. There are $32$ possibilities as a total... THX. – zoli Sep 27 '15 at 12:00
• @zoli Thanks, Can yo please check this question as well math.stackexchange.com/questions/1452840/… – Kittu Sep 27 '15 at 12:41
• @Kittu: Please hit the check mark if you liked the answer. – zoli Sep 27 '15 at 12:42