please show me how to do all parts of this question in detail--using general pro

Question

please show me how to do all parts of this question in detail--using general probability if applicable and also permutations/combinations/counting principle. thanks

Suppoue thaet you and your best friend (we'll call this your BFF in the rest of this problem) go out to a movie with & other friends Somehow you all get seated at random in a row of 6 seats a Whet is the probability that you and your BFF end up sitting next to each other? b. if you end up sitting at one of the two ends, then what is the probability that your BFF is sitting next to you? C it you end up sitting at one of the "interior" seats, then what is the probability that your BFF is sitting next to you d Consider your results for parts (a). (b), and (c). What should be the relationship between results? is it true for your results? e How many friends would you need to go out with before the probability of a random seating next to r significant other dropped below 10%? 5%?

Explanation / Answer

There are 6 friends.

Total number of ways they can be seated = 6! = 720.

(A)

Number of ways in which 2 people ( me and my BFF) sit together

= 5! ×2! = 120×2=240

[Since, assuming they sit together, there effectively be 5 people to be arranged, so 5!. Now, the two people ( me and my BFF) can interchange places, which is possible in 2! Ways.]

So the required probability = 240/720 = 1/3

(B)

if I end up sitting in one of the ends, i can sit in 2P1 = 2 Ways since there are two ends.

Now 5 spaces will be remaining and my BFF must sit next to me and he has only one way to do that.

Since 2 of the 6 places are now occupied, the remaining 4 people will arrange themselves in the remaining chairs in 4! Ways .

So the total number of favourable ways = 2P1 × 4! = 2× 24 =48

And the required probability = 48/720 = 1/15

(C)

Since I seat in an interior seats, that can happen in 4P1 = 4 ways.

Now, given that I'm seated, my BFF will have 2P1 = 2 ways to sit (that is, either to my left or to my right)

Now that the two of us are seated, the other 4 friends can arrange themselves in 4! = 24 ways  

So the total number of favourable ways = 4P1 × 2P1 × 4! = 4×2×24 = 192

And the required probability = 192 / 720 = 4/15

(D)

Considering parts (A) , (B) & (C)

Part (A) = part (B) + part (C)

probability that we sit together provided I sit in one of the extremes + probability that we sit together provided i sit in the interior

= 1/15 + 4/15 = 5/15 = 1/3

This is equal to the probability that we sit next to each other (that is part (A)).

(E)

Let X friends go out together.

Let the event that particular 2 of them sit together be denoted by G

We know that P(G) < 10%

Or, P(G) < 10/100 . Or, P(G) < 1/10.

P(G) = (x-1)! × 2! / x!

Since, let's consider that the two people who want to sit together as one person. So the total number of people to be seated = x-1. And the number of ways that this can happen = (x-1)!

Now, the two people who were considered as a single person can change places amongst themselves in 2! Ways.

And the total number of ways in which all of them can be arranged = x!

So, the problem becomes to find x such that :

(x-1)! × 2 /x! < 1/10

Or, (x-1)! × 2 / ( x × (x-1)!) <1/10

Or, 2/x <1/10

Or, x>20

Thus there should be a total of 20 people for the required probability to be achieved. And thus one needs to go out with 19 friends.

Similarly,

For P(G) < 5%

Or, P(G) < 5/100. Or, P(G) < 1/20

Or (x-1)!×2 /X! < 1/20

Or, 2/x< 1/20

Or, x> 20×2

Or x> 40.

Thus, one needs to go out with 39 friends before the probability of a random seating next to their significant other drops below 5 %.

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