Question

) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.

Answer 1

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= $\dfrac{1}{2}$

The length of bits : n = 10

Let X = Number of getting ones.

Then , $X \sim Bin(n=10,\ p=\dfrac{1}{2})$

Binomial distribution formula : $P(X=x)=^nC_x p^x q^{n-x}$ , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , $p=q=\dfrac{1}{2}$

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

$P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}$

$=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}$

$=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})$

$=(\dfrac{1}{2})^{10}(210+252)$

$=(0.0009765625)(462)$

$=0.451171875\approx0.4512$

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.