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 November 14th, 2012, 06:58 AM #1 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 probability 1. A firm plans to bid Rs.300 per tonne for a contract to supply 1000 tonnes of metal. It has 2 competitors A and B and it assumes that the probability that A will bid less than Rs.300 per tonne is 0.3 and that B will bid less than Rs.300 per tonne is 0.7. If the lowest bidder gets all the business and the firms bid independently what is the expected value of the contract to the firm? 2. A pitcher is taken to the well every day for 4 years. If the odds be 100:1 against its being broken on any particular day, show that the chance of its ultimately surviving is more than 1/4. 3. A manager has 2 assistants and he bases his decision on information supplied independently by each of them. The probability that he makes a mistake in his thinking is 0.005. The probability that an assistant gives wrong information is 0.3. Assuming that the mistakes made by the manager are independent of the information given by the assistants, find the probabity that he reaches a wrong decision. (ans. 0.513) Please explain the solutions in detail.
 November 14th, 2012, 09:01 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,654 Thanks: 2087 (1).  The 1000 tonnes of metal would fetch Rs. 300000, so multiply that amount by (1 - 0.3)(1 - 0.7), the probability that both of the competitors fail to underbid the firm. Are you sure that Q2 is typed correctly? A Google search suggests that the odds quoted should be 1000:1 and that "more than" should be "rather less than". Are you sure that Q3 is typed correctly? A Google search suggests the sentence "The probability that he makes a mistake in his thinking is 0.005." is missing and that the intended answer is 0.51245.
November 14th, 2012, 10:42 PM   #3
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 Originally Posted by skipjack (1).  The 1000 tonnes of metal would fetch Rs. 300000, so multiply that amount by (1 - 0.3)(1 - 0.7), the probability that both of the competitors fail to underbid the firm. Are you sure that Q2 is typed correctly? A Google search suggests that the odds quoted should be 1000:1 and that "more than" should be "rather less than". Are you sure that Q3 is typed correctly? A Google search suggests the sentence "The probability that he makes a mistake in his thinking is 0.005." is missing and that the intended answer is 0.51245.
I've typed Q2 correctly, but Q3 incorrectly(I've edited it). I'm extremely sorry! Please tell me how to solve these 2 problems. In Q1 I don't understand who is supplying metals to whom. Could you please tell me?

 November 15th, 2012, 10:25 AM #4 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: probability For Q2, assuming the numbers that you gave are correct and assuming that there are 1461 days in 4 years, I get the following: The probability of the pitcher being broken is the probability of the pitcher being broken on the first day or (plus) the probability of the pitcher being broken on the second day, etc. So it is a summing function and it is given by: $\frac{1}{100}+\frac{99}{100}\cdot \frac{1}{100}+(\frac{99}{100})^2\cdot \frac{1}{100}+...+(\frac{99}{100})^{1460}\cdot \frac{1}{100}$ $=\frac{1}{100}\left(1+\frac{99}{100}+(\frac{99}{10 0})^2+...+(\frac{99}{100})^{1460}\right)$ $=\frac{1}{100} \cdot \frac{\frac{99}{100}^{1461}-1}{\frac{99}{100}-1}$ = 0.99999958022235476469642315307448 (again, this is the probability of it being broken). The probability of the pitcher surviving is the probability of the pitcher surviving on the first day and the probability of the pitcher surviving on the second day, etc. So it is a multiplicative function and it is given by: $(\frac{99}{100})^{1461}$ = 4.1977764523530357684692552352471e-7 (the two numbers sum to 1). --------------------------------------------- For the numbers that [color=#00BF00]skipjack[/color] gave, I get: $=\frac{1}{1000} \cdot \frac{\frac{999}{1000}^{1461}-1}{\frac{999}{1000}-1}$ = 0.76816537560150533002164851762633 (this is the probability of it being broken). The probability of the pitcher surviving is: $(\frac{999}{1000})^{1461}$ = 0.23183462439849466997835148237367 (the two numbers sum to 1)
 November 15th, 2012, 07:32 PM #5 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: probability Q3 is, in my opinion, very poorly stated (I would like to know what others think about it and how they interpret it). I will get to my objections in a bit, but first things first: The probability that the 2 assistants will both be right is $\ (1-0.3)^2=0.7^2=0.49 \$, the probability that the 2 assistants will both be wrong is $\ 0.3^2=0.09 \$ and the probability that one will be wrong and one will be right is $\ 2*0.3*0.7=0.42$ ------------------------- The answer that I get that is closest to the given answer (0.513) is if the question reads "For all decisions, in a highly multi-valued system (one right choice, and many, many possible wrong choices), what is the probability that all agree and the right decision is not made?" I work the problem as follows: $P=1.0-0.7^2(0.995)=0.51245 \$ where the value between the '-' sign and the '=' sign is the probability that the right decision is made. --------------------------- The answer that I get that is next closest to the given answer is if the question reads "For all decisions, in a two valued system (one right choice, one wrong choice), what is the probability that all agree (including mistakes) and the right decision is not made?" I work the problem as follows: $P=1.0-\left(0.7^2(0.995)+0.3^2(0.005)\right)=0.512 \$ where the value between the '-' sign and the '=' sign is the probability that the right decision is made. -------------------------- The wording above avoided the decision making process when the assistants do not agree, in which case one is right, and the other is wrong. What happens then? Who knows? But let's try. We can assume that half the time the manager decides to go with the assistant that is correct and half the time he decides to go with the assistant that is wrong. So now we can word the problem "What is the probability that the manager makes the wrong decision". First, the multi-valued system. The calculation is as follows: $P=1.0-\left(0.7^2(0.995)+\frac{2*0.3*0.7}{2}(0.995)\righ t)=0.3035 \$ where the value between the '-' sign and the '=' sign is the probability that the right decision has been made. ---------------------------- For the two-valued system: $P=1.0-\left(0.7^2(0.995)+0.3^2(0.005)+ \frac{2*0.3*0.7}{2}(0.995+0.005)\right)=0.302 \$ where the value between the '-' sign and the '=' sign is the probability the right decision has been made. ------------------------- Finally, let's say the manager does not make a decision unless both assistants agree. And let's just assume a two valued system. Out of all decisions to be made, the probability of making a right decision is: $P_{right}_{(all)}=0.7^2(0.995)+0.3^2(0.005)=0.488 \$ and the probability of a wrong decision is: $P_{wrong}_{(all)}=0.7^2(0.005)+0.3^2(0.995)=0.092 \$ Since no decision is just that: no decision, the probability of a wrong decision (counting only decisions that were actually made) is: $\frac{0.092}{0.092+0.488}=\frac{0.092}{1-2*0.3*0.7}=0.1586$ Again, in my opinion, a very poorly stated problem.
 November 15th, 2012, 11:28 PM #6 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: probability OK! I'm able to solve Q2 and Q3 now. Thanks a lot both of you. But please tell me who is supplying metals to whom in Q1. Is it the firm who is supplier and A &B are receivers? Please please tell me...
 November 16th, 2012, 06:17 PM #7 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: probability In Q1, the firm is bidding against A and B in order to supply the metal to an unnamed receiver. The lowest bidder gets to be the supplier. Hopefully, the solution provided by [color=#00BF00]skipjack[/color] is now clear.
November 17th, 2012, 04:35 AM   #8
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Quote:
 Originally Posted by jks In Q1, the firm is bidding against A and B in order to supply the metal to an unnamed receiver. The lowest bidder gets to be the supplier. Hopefully, the solution provided by [color=#00BF00]skipjack[/color] is now clear.
I got it now. Thank you!

 May 2nd, 2018, 04:49 PM #9 Newbie   Joined: May 2018 From: Nepal Posts: 1 Thanks: 0 why the images are not visible in q. no. 3?
May 2nd, 2018, 05:43 PM   #10
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 Originally Posted by iamashma why the images are not visible in q. no. 3?
because the thread is 10 years old and the server they were on is probably in a junkyard somewhere.

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