3. Suppose Mr. and Mrs. Smith along with 8 other individuals will be randomly se

Question

3. Suppose Mr. and Mrs. Smith along with 8 other individuals will be randomly seated in 10 chairs that are arranged in a row in a concert hall. Find the probability that Mr. and Mrs Smith will end up seating next to each other.

4. The system shown below is a segment of an electrical circuit which has three relays. The current will flow from point (a) topoint (b), that is system is working, if there is at least one close path when the relays are switched to "close". However, the relays may malfunction. Suppose when the switch is thrown, the probability that relay 1 closes properly is .3, relay 2 closes properly is .5 and relay 3 closes properly is .8, where relays are working independent of each others.

system

---------------(1)--------------

a*--------| |---------------(3)-----------*b

----------------(2)--------------   

a. Find the probability that system is working (that is, current is flowing from point a to point b). That is find P(W), where W denotes the event that system is working.

b. Find the (conditional) probability that the component 1 is not working given that we know that system is working. That is find P(A1| W)

5. In a computer lab, 30% of the microcomputers are IBM, 50% are Apple, and the rest are Dell microcomputers. Suppose that at any given time the probability that a microcomputer works properly is 92% if it is IBM; 72% if it is Apple and 98% if it is a Dell. One computer is selected at random from this lab. Let W be the event that selected computer is working. Find theconditional probability that the selected computer is a Dell computer given that it is working properly. That is find P(Dell | W).

Explanation / Answer

3.If we want the couple to sit together, we treat them as 1 body (but can be permuted in 2! ways). Thus, the other 8 has 8! ways to permute.

Thus, we have 2!8! ways to do the job.

Thus,

P = (2!8!)/10! = 0.0222

4.

P( relay 1 closes) = 0.3
P( relay 2 closes) = 0.5
P( relay 3 closes) = 0.8

a)
P(W) = P( System is working fine) = P(A1)*P(A2)*P(A3) = 0.3*0.5*0.8 = 0.12

b)
P(A'1/W} = P( A'1 n W) / P(W)
P(A'1) = 1 - P(A) = 0.7
P(A'1 n W) = P(A1)*P(W) = 0.7*0.12 = 0.084

P(A'1/W} = P( A'1 n W) / P(W) = 0.084 / 0.12 = 0.7

5. the probabilities of the computer brands add up to 100%, then P(Dell) = 0.20.

Let

I = IBM
A = Apple
D = Dell
W = working

Thus,

P(D|W) = P(D n W) / P(W)

As

P(D n W) = P(D) P(W|D) = 0.20*0.98 = 0.196

and

P(W) = P(I) P(W|I) + P(A) P(W|A) + P(D) P(W|D)

= 0.30*0.92 + 0.50*0.72 + 0.20*0.98

P(W) = 0.832

Thus,

P(D|W) = P(D n W) / P(W) = 0.235576

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