Answer :
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) Two computers are chosen at random.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
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
LET X = Number of computers that are classified as ancient
So, it means X ~ [tex]Binom(n=2, p=0.94)[/tex]
Now, Probability that both computers are ancient is given by = P(X = 2)
P(X = 2) = [tex]\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}[/tex]
= [tex]1 \times 0.94^{2} \times 1[/tex]
= 0.8836
(b) Eight computers are chosen at random.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
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
LET X = Number of computers that are classified as ancient
So, it means X ~ [tex]Binom(n=8, p=0.94)[/tex]
Now, Probability that all eight computers are ancient is given by = P(X = 8)
P(X = 8) = [tex]\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}[/tex]
= [tex]1 \times 0.94^{8} \times 1[/tex]
= 0.6096
(c) Here, also 8 computers are chosen at random.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
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
LET X = Number of computers classified as cutting dash edge
So, it means X ~ [tex]Binom(n=8, p=0.06)[/tex]
Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X [tex]\geq[/tex] 1)
P(X [tex]\geq[/tex] 1) = 1 - P(X = 0)
= [tex]1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}[/tex]
= [tex]1 - [1 \times 1 \times 0.94^{8}][/tex]
= 1 - [tex]0.94^{8}[/tex] = 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.