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Telephone calls

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by MURALIDHARGAJULA » Fri Jun 03, 2011 12:07 am
A company has 8 telephone lines. IF 4 calls have to be made by the company , what is the probability that they are each made on a different line?

A) 35/2048
B) 105/256
C) 105/2048
D) 35/256
E) 105/128

[spoiler]OA: B[/spoiler]
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by cans » Fri Jun 03, 2011 12:16 am
total possible ways = 8^4 (each call can be made on any of the 8 lines)
if different lines, select 4 lines and arrange them - 8C4 * 4!
thus prob = 8C4 * 4! / 8^4 = 8*7*5*6/(8*8*8*8) = 105/256
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by Anurag@Gurome » Fri Jun 03, 2011 12:24 am
MURALIDHARGAJULA wrote:A company has 8 telephone lines. IF 4 calls have to be made by the company , what is the probability that they are each made on a different line?
Total number of ways to make 4 calls using 8 lines = 8^4

But we have to make calls on different lines.
Hence, we have to select 4 lines out of 8 and distribute the 4 calls in those selected lines. Which can be done in (8C4)*4! ways

Therefore, required probability = (8C4)*4!/(8^4) = 8*7*5*6/8*8*8*8 = 105/256

The correct answer is B.
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by ganesh prasath » Fri Jun 03, 2011 5:21 am
"Hence, we have to select 4 lines out of 8 and distribute the 4 calls in those selected lines. Which can be done in (8C4)*4! ways"

I didnt understand this part could u please explain this to me..
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by Anurag@Gurome » Fri Jun 03, 2011 6:00 am
ganesh prasath wrote:"Hence, we have to select 4 lines out of 8 and distribute the 4 calls in those selected lines. Which can be done in (8C4)*4! ways"

I didnt understand this part could u please explain this to me..
The company has 8 lines and they have to make 4 calls each on different line. Hence they need 4 different lines to do so. Hence, we are selecting 4 lines out of 8 lines in 8C4 ways.

Now we have to assign the 4 calls to these 4 selected lines. Which can be done in 4! ways, because the first call can be made in any of the 4 selected lines. The second call can be made in any of the remaining 3 selected lines. The third call can be made in any of the remaining 2 selected lines. And the fourth and last call can be made in the remaining selected line.

Hence, we can assign the 4 calls to these 4 selected lines in 4*3*2*1 = 4! ways

Hope it helps.
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by vikram4689 » Fri Jun 03, 2011 6:32 am
There is one easier way:

Selection of 4 different telephone lines : 8*7*6*5
Selection of 4 telephone lines (no restriction) : 8*8*8*8

Prob. : (8*7*6*5)/(8*8*8*8) = 105/256
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by GMATGuruNY » Fri Jun 03, 2011 7:00 am
MURALIDHARGAJULA wrote:A company has 8 telephone lines. IF 4 calls have to be made by the company , what is the probability that they are each made on a different line?

A) 35/2048
B) 105/256
C) 105/2048
D) 35/256
E) 105/128

[spoiler]OA: B[/spoiler]
Another approach:

The first call is irrelevant: it can be placed on any of the 8 lines.

P(2nd call uses a different line from the 1st) = 7/8.
P(3rd call uses a different line from the 1st and 2nd) = 6/8 = 3/4.
P(4th call uses a different line from the 1st, 2nd and 3rd) = 5/8.

Since we want all of the events above to happen together, we multiply the results:
P(all 4 calls use a different line) = 7/8 * 3/4 * 5/8 = 105/256.

The correct answer is C.
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by aftableo2006 » Sun Jun 05, 2011 11:38 pm
B is the answer
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