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)
P(X
1) = 1 - P(X = 0)
= 
=
= 1 -
= 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.