Binary Probability help

vertex

New member
Joined
Apr 17, 2021
Messages
5
Suppose we wanted to create a 12-digit bit string (a bit string is a binary number consisting of only ones and zeros).

How many different 12-digit bit strings are possible?
For this one I did \(\displaystyle 2^{12}\) \(\displaystyle = 4096\) Different strings

What is the probability that a12-digit bit string has exactly five ones but begins and ends in zero?(Hint: Remember, all of the different 1’s and 0’s are indistinguishable from one another.)

What is the probability that a12-digit bit string has at least eight ones but begins and ends in zero? (Hint: break it up into cases)

The last 2 have me a bit puzzled as I am not sure where to begin. Could someone tutor me.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
10,907
[QUOTE="vertex, post: 535246, member: 92038"
What is the probability that a12-digit bit string has at least eight ones but begins and ends in zero? (Hint: break it up into cases)[/QUOTE]
Lets play the back-of-the-book game. The textbook has the answers. So we look the answer up.
We see \(\displaystyle\sum\limits_{k = 8}^{10} {\frac{{10!}}{{k!(10 - k)!}}} \), What the heck?
Well you explain it!
 
  • Like
Reactions: lex

lex

Full Member
Joined
Mar 3, 2021
Messages
387
We see \(\displaystyle\sum\limits_{k = 8}^{10} {\frac{{10!}}{{k!(10 - k)!}}} \), What the heck?
Well you explain it!
Probability
 

vertex

New member
Joined
Apr 17, 2021
Messages
5
Would that be \(\displaystyle 12!\) / \(\displaystyle 8!(12-8)!\) = \(\displaystyle 12!\) / \(\displaystyle 8!4!\) = \(\displaystyle 12*11*10*9\) / \(\displaystyle 4*3*2*1\) = \(\displaystyle 11880 / 24\) = \(\displaystyle 49.17\)
 

lex

Full Member
Joined
Mar 3, 2021
Messages
387
What is the probability that a12-digit bit string has exactly five ones but begins and ends in zero?(Hint: Remember, all of the different 1’s and 0’s are indistinguishable from one another.)
That effectively means you have 10 spaces into which you can put five ones. Work out how many arrangements there are for this and divide by 4096 to get the probability.
(Assuming all 4096 strings are equally likely to occur).
 

lex

Full Member
Joined
Mar 3, 2021
Messages
387
Your post appeared after I made my post. Do you know how many combinations there are when choosing 5 objects from 10?
 
Top