This problem has been taken from old paper of 'Master of Business Administration examination' of university in India. Problem goes as follows: A certain make of computer has a probability of 0.05 of giving wrong results owing to electrical or mechanical failure. To improve the accuracy of calculations the operator feeds the same data into five such statistically independent computers and accept as accurate the answer that majority of the computers give. For a calculation, what is the probability that a majority of the computers are operating properly?Answer: Majority of computers means more than 3. i-e 4 or 5 computers. So as per my calculations the probability that 4 or 5 computers are operating properly is 0.9774075.Is this answer correct? Let me know from this forum.Btw, correct answer is not available with me.

June 10th 2012, 08:56 AM

Plato

Re: binomial probability problem

Quote:

Originally Posted by Vinod

This problem has been taken from old paper of 'Master of Business Administration examination' of university in India. Problem goes as follows: A certain make of computer has a probability of 0.05 of giving wrong results owing to electrical or mechanical failure. To improve the accuracy of calculations the operator feeds the same data into five such statistically independent computers and accept as accurate the answer that majority of the computers give. For a calculation, what is the probability that a majority of the computers are operating properly?Answer: Majority of computers means more than 3. i-e 4 or 5 computers. So as per my calculations the probability that 4 or 5 computers are operating properly is 0.9774075.Is this answer correct? Let me know from this forum.Btw, correct answer is not available with me.

If that is what it means, four or five, then yes that is correct.

June 10th 2012, 09:50 AM

Wilmer

Re: binomial probability problem

OK you probabilitizers:
21 cards are labelled: 1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,7,8,9,10,11
You pick 5 cards at random, no replacement.
To "win", you need at least 3 DIFFERENT cards < 6.
Example: two 3's counts as one 3; three 4's counts as one 4.
So, as example, 5,2,5,9,5 is not a winning combo...got that?
But 3,1,3,3,5 is a winner...ok?
What's the probability of picking a winning combo?

NOTE: couldn't open a "new post"; so used this one: thanks Vinod!

June 11th 2012, 07:08 AM

Vinod

Re: binomial probability problem

Thanks for your reply. Your problem is interesting and little bit difficult also. However, I shall give you answer very soon.okay.

June 11th 2012, 08:09 AM

HallsofIvy

Re: binomial probability problem

Quote:

Originally Posted by Vinod

This problem has been taken from old paper of 'Master of Business Administration examination' of university in India. Problem goes as follows: A certain make of computer has a probability of 0.05 of giving wrong results owing to electrical or mechanical failure. To improve the accuracy of calculations the operator feeds the same data into five such statistically independent computers and accept as accurate the answer that majority of the computers give. For a calculation, what is the probability that a majority of the computers are operating properly?Answer: Majority of computers means more than 3. i-e 4 or 5 computers.

Plato responded "If that is what it means, four or five, then yes that is correct." But, in fact, with 5 computers, 3 giving an specific response would mean that only 2 give another response and 3> 2. To me, "majority" would mean "3 or more", not "more than 3". Where did you get that?

Quote:

So as per my calculations the probability that 4 or 5 computers are operating properly is 0.9774075.Is this answer correct? Let me know from this forum.Btw, correct answer is not available with me.

The probability that exactly 3 operate correctly is . Add that to the probability that "4 or 5 are operating correctly" to get the probability that "more than half are operating correctly".

June 12th 2012, 06:11 AM

Vinod

Re: binomial probability problem

Quote:

Originally Posted by Wilmer

OK you probabilitizers:21 cards are labelled: 1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,7,8,9,10,11You pick 5 cards at random, no replacement.To "win", you need at least 3 DIFFERENT cards < 6.Example: two 3's counts as one 3; three 4's counts as one 4.So, as example, 5,2,5,9,5 is not a winning combo...got that?But 3,1,3,3,5 is a winner...ok?What's the probability of picking a winning combo? NOTE: couldn't open a "new post"; so used this one: thanks Vinod!

Yesterday, I replied you that I would answer your problem very soon.Here is the answer. The probability of picking a winning combo i-e at least 3 DIFFERENT cards is .

June 12th 2012, 06:47 AM

Vinod

Re: binomial probability problem

Quote:

Originally Posted by HallsofIvy

Plato responded "If that is what it means, four or five, then yes that is correct." But, in fact, with 5 computers, 3 giving an specific response would mean that only 2 give another response and 3> 2. To me, "majority" would mean "3 or more", not "more than 3". Where did you get that? The probability that exactly 3 operate correctly is . Add that to the probability that "4 or 5 are operating correctly" to get the probability that "more than half are operating correctly".

"The median is that value of the variable which divides the group in two equal parts, one part comprising all the values greater and the other, all the values less than median" In this 5 group of computers, the median value is 3. I considered here median as arithmetic mean.because actual arithmetic mean is 2.5. So according to my point of view, majority means greater than median. So i considered only probability of more than 3 computers operating properly and not more than 2.5 computers operating properly. However, thanks for putting forward your point of view.

June 12th 2012, 07:00 AM

Wilmer

Re: binomial probability problem

Quote:

Originally Posted by Vinod

The probability of picking a winning combo i-e at least 3 DIFFERENT cards is .

No. You are way too low.
If you were allowed only 3 cards instead of 5, the probability is: 15/21 * 12/18 * 9/19 = 27/133
Being allowed 5 cards as is the case in this problem makes that even higher.

June 12th 2012, 11:28 AM

Wilmer

Re: binomial probability problem

Here's the solution:

Spoiler:

Relabel the cards: 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c 6 7 8 9 10 11
so there are C(21, 5) equally likely hands.

Sort the hand. Then a winning combination must be one of the following.

A)

Two high cards and three different low cards
C(6,2) * C(5,3) * 3 * 3 * 3 = 150 * 27 = 4050
(The last factors of 3 are the selection of a, b, or c.)

B)

One high card, three different low cards, and a matching low card
C(6,1) * C(5,3) * 3 * 3 * 3 * 3 * 2 / 2 = 60 * 81 = 4860
(The division by 2 is to allow for the order of the match.)
One high card and four different low cards
C(6,1) * C(5,4) * 3 * 3 * 3 * 3 = 2430

C)

Two pairs of matching low cards, and a different low card
C(5,2) * 3 * 3 * C(3,1) * 3 = 810
(The middle factors of 3 are the omission of a, b, or c.)
A triple low card, and two different low cards
C(5,1) * C(4,2) * 3 * 3 = 270
Four different low cards, and a matching low card
C(5,4) * 3 * 3 * 3 * 3 * 4 * 2 / 2 = 1620
Five different low cards
3 * 3 * 3 * 3 * 3 = 243

So the probability of a winning combo is
(4050 + 4860 + 2430 + 810 + 270 + 1620 + 243) / C(21,5)
= 14283 / 20349
= 1587 / 2261
= 70.2 %

June 12th 2012, 12:46 PM

HallsofIvy

Re: binomial probability problem

Quote:

Originally Posted by Vinod

"The median is that value of the variable which divides the group in two equal parts, one part comprising all the values greater and the other, all the values less than median" In this 5 group of computers, the median value is 3. I considered here median as arithmetic mean.because actual arithmetic mean is 2.5. So according to my point of view, majority means greater than median. So i considered only probability of more than 3 computers operating properly and not more than 2.5 computers operating properly. However, thanks for putting forward your point of view.

Well, my point of view seems to be the one you are quoting. I just do not understand why, having said yourself that the median is 2.5, you then choose to ignore that.

June 15th 2012, 06:31 AM

Vinod

Re: binomial probability problem

Quote:

Originally Posted by Wilmer

Here's the solution:

Spoiler:

Relabel the cards: 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c 6 7 8 9 10 11so there are C(21, 5) equally likely hands.Sort the hand. Then a winning combination must be one of the following.A)Two high cards and three different low cardsC(6,2) * C(5,3) * 3 * 3 * 3 = 150 * 27 = 4050(The last factors of 3 are the selection of a, b, or c.)B)One high card, three different low cards, and a matching low cardC(6,1) * C(5,3) * 3 * 3 * 3 * 3 * 2 / 2 = 60 * 81 = 4860(The division by 2 is to allow for the order of the match.)One high card and four different low cardsC(6,1) * C(5,4) * 3 * 3 * 3 * 3 = 2430C)Two pairs of matching low cards, and a different low cardC(5,2) * 3 * 3 * C(3,1) * 3 = 810(The middle factors of 3 are the omission of a, b, or c.)A triple low card, and two different low cardsC(5,1) * C(4,2) * 3 * 3 = 270Four different low cards, and a matching low cardC(5,4) * 3 * 3 * 3 * 3 * 4 * 2 / 2 = 1620Five different low cards3 * 3 * 3 * 3 * 3 = 243So the probability of a winning combo is(4050 + 4860 + 2430 + 810 + 270 + 1620 + 243) / C(21,5)= 14283 / 20349= 1587 / 2261= 70.2 %

Your answer is correct. I solved it wrongly. Okay.

June 15th 2012, 06:48 AM

Vinod

Re: binomial probability problem

Quote:

Originally Posted by HallsofIvy

Well, my point of view seems to be the one you are quoting. I just do not understand why, having said yourself that the median is 2.5, you then choose to ignore that.

If the number of observations is odd, then the median is the middle value after the observations have been arranged in ascending or descending order of magnitude. In this question number of computers is 5 which is odd.So the median is 3. Majority means greater than median.Hence my answer contains only the probability of 4 or 5 computers operating properly.