Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be :
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are : 
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Answer: 1) 0.10
2) 0.60
3) 0.20
4) 0.10
<u>Step-by-step explanation:</u>
The total frequency is 20+120+40+20 = 200. This means they ran the experiment 200 times. The probability distribution is calculated by the satisfactory number of outcomes (frequency) divided by the total number of experiments/outcomes (total frequency):
![\begin{array}{c|c||lc}\underline{x}&\underline{f}&\underline{f\div 200}&\underline{\text{Probability Distribution}}\\1&20&20\div200=&0.10\\2&120&120\div 200=&0.60\\3&40&40\div 200=&0.20\\4&20&20\div 200=&0.10\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Cc%7C%7Clc%7D%5Cunderline%7Bx%7D%26%5Cunderline%7Bf%7D%26%5Cunderline%7Bf%5Cdiv%20200%7D%26%5Cunderline%7B%5Ctext%7BProbability%20Distribution%7D%7D%5C%5C1%2620%2620%5Cdiv200%3D%260.10%5C%5C2%26120%26120%5Cdiv%20200%3D%260.60%5C%5C3%2640%2640%5Cdiv%20200%3D%260.20%5C%5C4%2620%2620%5Cdiv%20200%3D%260.10%5Cend%7Barray%7D%5Cright%5D)
Y=3x-2
y=3(4)-2
y=12-2
y=10
I hope this helps.
Please ensure that you've copied down this problem correctly. You speak of someone named "Yarin" but also speak of "my" and "I". Are there really two different people who "star" in this problem? Are you studying "ratios" right now?