Answer:
(a) 0.8836
(b) 0.6096
(c) 0.3904
Step-by-step explanation:
We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.
(a) <u>Two computers are chosen at random.</u>
The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 2 computers
             r = number of success = both 2
            p = probability of success which in our question is % of computers
                   that are classified as ancient, i.e; 0.94
<em>LET X = Number of computers that are classified as ancient</em>
So, it means X ~ 
Now, Probability that both computers are ancient is given by = P(X = 2)
        P(X = 2)  = 
                       =  
 
                       = 0.8836
(b) <u>Eight computers are chosen at random.</u>
The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 8 computers
             r = number of success = all 8
            p = probability of success which in our question is % of computers
                   that are classified as ancient, i.e; 0.94
<em>LET X = Number of computers that are classified as ancient</em>
So, it means X ~ 
Now, Probability that all eight computers are ancient is given by = P(X = 8)
        P(X = 8)  = 
                       =  
 
                       = 0.6096
(c) <u>Here, also 8 computers are chosen at random.</u>
The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 8 computers
             r = number of success = at least one
            p = probability of success which is now the % of computers
                   that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06
<em>LET X = Number of computers classified as cutting dash edge</em>
So, it means X ~ 
Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X  1)
 1)
        P(X  1)  = 1 - P(X = 0)
 1)  = 1 - P(X = 0)
                       =  
                       = ![1 - [1 \times 1 \times 0.94^{8}]](https://tex.z-dn.net/?f=1%20-%20%5B1%20%5Ctimes%201%20%5Ctimes%200.94%5E%7B8%7D%5D) 
 
                       = 1 -  = 0.3904
 = 0.3904 
Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.
For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.
So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.