Question

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. What is the probability that both computers are ancient​? ​(b) Eight computers are chosen at random. What is the probability that all eight computers are ancient​? ​(c) What is the probability that at least one of eight randomly selected computers is cutting dash edge​? Would it be unusual that at least one of eight randomly selected computers is cutting dash edge​? ​(a) Two computers are chosen at random. What is the probability that both computers are ancient​?

Answer 1

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;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

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 ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

       P(X = 2)  = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

                      = $1 \times 0.94^{2} \times 1$

                      = 0.8836

(b) Eight computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

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 ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

       P(X = 8)  = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

                      = $1 \times 0.94^{8} \times 1$

                      = 0.6096

(c) Here, also 8 computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

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 ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

       P(X $\geq$ 1)  = 1 - P(X = 0)

                      =  $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

                      = $1 - [1 \times 1 \times 0.94^{8}]$

                      = 1 - $0.94^{8}$ = 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.