a) There are 7 identical bags. In each bag, there are 9 balls, numbered from 1 to 9, inclusive. We randomly pick a ball from each bag. Determine the probability that product of the 7 number on the 7 balls is a multiple of 3. You do not need to simplify your answer.

b) Assuming 10⁵ bits are independently transmitted over a network in which the probability of an erroneous bit is 10^-5, calculate the probability of bit error when the total number of errors is less than or equal to 5. You do not need to simplify your answer.  

a. P = (1/3)⁷

b. P(X = 5) = (10⁵C5)(10^-5)⁵((1 - 10^-5)^(10⁵ - 5))

Step-by-step explanation

Let's interpret the given:

7 bags means there are 7 trials

9 balls, numbered 1 - 9 means 9 choices per bag

Condition: product of the 7 numbers is a multiple of 3

when is the product a multiple of 3

when either 3, 6 or 9 is drawn

So per bag:

that means 3 out of 9 choices

or P = 3/9 = 1/3

Given 7 bags/trials:

P = (1/3)⁷

b. Given:

n = 10⁵ (this is the total amount of bits)

P(X = 1) = 10^-5 (the probability of one error bit)

Problem: P(X =< 5) = ? (probability that the total number of errors is less than equal to 5)

This is a binomial distribution with the formula:

P(X = x) = nCx p^x (1-p)^{n - x}

where n is the total number of trials

p is the probability of one success (in this case error)

x = errors counted

C just means you get the combination or the number of ways of choosing x given n

Substituting the given:

P(X = 5) = (10⁵C5)(10^-5)⁵((1 - 10^-5)^(10⁵ - 5))

It's not asking yo to simplify because 10^5 is very big