Answer:
Step-by-step explanation:
From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .
So; as each place is fitted with either 0 or 1
[tex]\mathbf{|R|= 2^{10}}[/tex]
Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0
Now;
if a 0 bit and a 1 bit are equally likely
The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;
[tex]\mathbf{P(E) = \dfrac{|E|}{|R|}}[/tex]
so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:
[tex]\mathbf{{|E|}=1 }[/tex] ; [tex]\mathbf{|R|= 2^{10}}[/tex]
[tex]\mathbf{P(E) = \dfrac{1}{2^{10}}}[/tex]
