GATE Overflow - Recent questions and answers in Probability
http://gateoverflow.in/qa/mathematics/probability
Powered by Question2AnswerAnswered: TIFR2015-A-1
http://gateoverflow.in/29156/tifr2015-a-1?show=137995#a137995
P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(6) = 1+P(6)<br />
<br />
=> P(2,4,6) + P(3,6) + P(1) + P(5) = 1+P(6)<br />
<br />
=> 1/2 + 1/3 + 1/6 + P(5) = 1 + P(6)<br />
<br />
=> P(5) = P(6)<br />
<br />
Also, P(6)<=1/6 and P(6)<=1/3. In the worst case, P(5)=P(6)<=1/3Probabilityhttp://gateoverflow.in/29156/tifr2015-a-1?show=137995#a137995Sat, 15 Jul 2017 11:24:17 +0000probability
http://gateoverflow.in/137763/probability
a lot consists of 12 good pencils, 6 with minor defects and 2 with major defects. A pencil is chosen at random. The probability that this pencil is not defective is?Probabilityhttp://gateoverflow.in/137763/probabilityFri, 14 Jul 2017 09:19:07 +0000probability
http://gateoverflow.in/137762/probability
a lot consists of good pencils, 6 with minor defects and 2 with major defects. A pencil is chosen at random. The probability that this pencil is not defective is?Probabilityhttp://gateoverflow.in/137762/probabilityFri, 14 Jul 2017 09:18:35 +0000Answered: probability
http://gateoverflow.in/137217/probability?show=137224#a137224
<p>There are 4 matches to be played, For at least 7 points, India should either win 7 points or 8 points.
<br>
<br>
P(winning 7 points) = win 2 points in 3 matches and win 1 point in one match.
<br>
=$4C3*0.5^{3}*0.05^{1}$ = 4*0.125*0.05 = 0.025
<br>
<br>
P(winning 8 pints) = $4C4*0.5^{4}$ =0.0625
<br>
P(at least 7) = P( 7 points ) + P( 8 points) = 0.025+0.0625 = 0.0875
<br>
</p>
<p><strong>Option (b)!</strong></p>Probabilityhttp://gateoverflow.in/137217/probability?show=137224#a137224Tue, 11 Jul 2017 10:01:31 +0000Answered: Doubt
http://gateoverflow.in/136941/doubt?show=136994#a136994
The mode is the number which appears most often. Since both 3 and 4 appear two times each, the mode is both 3 and 4.Probabilityhttp://gateoverflow.in/136941/doubt?show=136994#a136994Mon, 10 Jul 2017 02:34:00 +0000Answered: Probability posion distribtuion
http://gateoverflow.in/135554/probability-posion-distribtuion?show=135617#a135617
<p>Answer</p>
<p style="text-align:center"><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=10492760893903641131"></p>
<p> </p>Probabilityhttp://gateoverflow.in/135554/probability-posion-distribtuion?show=135617#a135617Sat, 01 Jul 2017 16:37:20 +0000Answered: Probability of dice
http://gateoverflow.in/131870/probability-of-dice?show=135606#a135606
$\left ( x_{1}+x_{2}+x_{3}+x_{4} \right )=20$<br />
<br />
We have to find coefficient of $\left [ x^{20} \right ]$<br />
<br />
$\left ( x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right )^{4}$<br />
<br />
$=\left [ x^{4} \right ]\left ( 1+x+x^{2}+x^{3}+x^{4}+x^{5}\right )^{4}$<br />
<br />
$=\frac{\left ( 1-x^{6} \right )^{4}}{\left ( 1-x \right )^{4}}$<br />
<br />
Now,<br />
<br />
$\left ( 1-x^{6} \right )^{4}=1-4x^{6}+6x^{12}-4x^{18}+...$<br />
<br />
$\left ( 1-x\right )^{4}=1+4x+\binom{5}{2}.x^{2}+\binom{6}{3}.x^{3}+....$<br />
<br />
$=\binom{19}{16}-4\times \binom{13}{10}+6\times \binom{7}{4}=35$Probabilityhttp://gateoverflow.in/131870/probability-of-dice?show=135606#a135606Sat, 01 Jul 2017 14:20:02 +0000Gate EE 17
http://gateoverflow.in/135069/gate-ee-17
Assume that in a traffic junction, the cycle of traffic signal lights is 2 minutes of green(vehicle does not stop) and 3 minutes of red (vehicle stops). Consider the arrival time of vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time in minutes for the vehicle at the junction is _________Probabilityhttp://gateoverflow.in/135069/gate-ee-17Wed, 28 Jun 2017 10:26:41 +0000[Gate 2006 EC] Probability
http://gateoverflow.in/134674/gate-2006-ec-probability
<p>Three companies, X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below.</p>
<table border="1" cellpadding="0" cellspacing="0" style="width:100%">
<tbody>
<tr>
<td> Company</td>
<td> % of computers supplied </td>
<td> Probability of being defective</td>
</tr>
<tr>
<td> X</td>
<td> 60%</td>
<td> 0.01</td>
</tr>
<tr>
<td> Y</td>
<td> 30%</td>
<td> 0.02</td>
</tr>
<tr>
<td> Z</td>
<td> 10%</td>
<td> 0.03</td>
</tr>
</tbody>
</table>
<p>Given that a computer is defective, the probability that it was supplied by Y is: [2 marks]
<br>
(A) 0.1
<br>
(B) 0.2
<br>
(C) 0.3
<br>
(D) 0.4</p>
<p>I got the correct answer by applying bayees theorem. But that theorem is applicable where the events are collectively exhaustive and mutual exclusive. These are collectively exhaustive but i don't think that they are mutually exclusive as Supplying of x does not mean that y wont supply.Please clear this doubt.</p>Probabilityhttp://gateoverflow.in/134674/gate-2006-ec-probabilitySun, 25 Jun 2017 06:07:49 +0000Answered: GATE2012_33
http://gateoverflow.in/1751/gate2012_33?show=134526#a134526
Total Probablity = 1/6 + 3/6{9/18} = 5/12Probabilityhttp://gateoverflow.in/1751/gate2012_33?show=134526#a134526Fri, 23 Jun 2017 17:18:21 +0000Answered: GATE2002-2.10
http://gateoverflow.in/840/gate2002-2-10?show=133892#a133892
<p>Do not know if it is a correct approach. Some one please verify.
<br>
Suppose array contains 10 elements.
<br>
'i' will be chosen at random, so probability that we will get our element in first try = 1/10.
<br>
Expected number of try for getting the element is 1/p = 10
<br>
<a rel="nofollow" href="http://gateoverflow.in/3561/gate2006-it-22">http://gateoverflow.in/3561/gate2006-it-22</a>
<br>
<br>
Hence answer is a) n .</p>Probabilityhttp://gateoverflow.in/840/gate2002-2-10?show=133892#a133892Mon, 19 Jun 2017 14:51:56 +0000Answered: GATE2002-2.16
http://gateoverflow.in/846/gate2002-2-16?show=133528#a133528
probability of getting head=p=1/2;<br />
<br />
by the formula p(x)=nCx*P^xq^(n-x)<br />
<br />
p(atleast one tail)=p(x>=1)==1-p(x<1)=4C0*(1/2)^4=1-1/16=15/16<br />
<br />
p(atleast one head)=p(x>=1)==1-p(x<1)=4C0*(1/2)^4=1-1/16=15/16<br />
<br />
so<br />
<br />
P(head&tail)=15/16*15/16=7/8Probabilityhttp://gateoverflow.in/846/gate2002-2-16?show=133528#a133528Sat, 17 Jun 2017 07:30:33 +0000Answered: Sheldon Ross chapter 2 questions 26
http://gateoverflow.in/132384/sheldon-ross-chapter-2-questions-26?show=133481#a133481
Nice question...the hint made me solve it...First better to find individual probabilities of sum of cubes roled, this one will be,<br />
<br />
2 : (1, 1);<br />
<br />
3 : (1, 2),(2, 1);<br />
<br />
4 : (1, 3),(2, 2),(3, 1);<br />
<br />
5 : (1, 4),(2, 3),(3, 2),(4, 1);<br />
<br />
6 : (1, 5),(2, 4),(3, 3),(4, 2),(5, 1);<br />
<br />
7 : (1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1);<br />
<br />
8 : (2, 6),(3, 5),(4, 4),(5, 3),(6, 2);<br />
<br />
9 : (3, 6),(4, 5),(5, 4),(6, 3);<br />
<br />
10 : (4, 6),(5, 5),(6, 4);<br />
<br />
11 : (5, 6),(6, 5);<br />
<br />
12 : (6, 6)<br />
<br />
Now direct chances of winning are 7 and 11 and P(Direct Win) = 8/36 and P(Direct loss) = 4/36 (i.e case 2,3 or 12). Now we need to consider cases (4,5,6,8,9,10) where we keep on playing until we get the same event before a 7.<br />
<br />
We know P(4) = 3/36 . Suppose 'n' be no of rolls made. So if I get 4 in the first roll next (n-2) rolls should be anything other than roll of 7 until we get 4 again.Let i is any of (4,5,6,8,9,10).<br />
<br />
P(Ei,n) = P(i)^2 (30 − Pi)^(n−2) /36^n .Now take summation from n = 2 to $infty$. you will get G.P series kind of sum.Ultimately you will get.<br />
<br />
P(i)^2 / 36(6 + P(i)) .Now add the probabilities for each case (i.e (4,5,6,8,9,10)) along with P(Direct win).<br />
<br />
<br />
<br />
Please pardon my writing, not yet well versed inserting symbols in answers.My answer coming as 0.49 almost.Probabilityhttp://gateoverflow.in/132384/sheldon-ross-chapter-2-questions-26?show=133481#a133481Fri, 16 Jun 2017 20:20:09 +0000Answered: Uniform probability distribution
http://gateoverflow.in/16426/uniform-probability-distribution?show=133404#a133404
<p style="text-align:center"><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=3610408301358408488"></p>
<p>Answer</p>Probabilityhttp://gateoverflow.in/16426/uniform-probability-distribution?show=133404#a133404Fri, 16 Jun 2017 12:28:09 +0000Answered: Continuous random distribution
http://gateoverflow.in/19949/continuous-random-distribution?show=133400#a133400
<p>Correct me if I am wrong</p>
<p style="text-align:center"><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=16724188865296904665"></p>
<p> </p>Probabilityhttp://gateoverflow.in/19949/continuous-random-distribution?show=133400#a133400Fri, 16 Jun 2017 12:07:17 +0000Answered: IISc PhD
http://gateoverflow.in/132635/iisc-phd?show=132672#a132672
I have solved it by trial and error method.<br />
<br />
Say 4 balls in the bag.<br />
<br />
3 balls are red and 1 is black.<br />
<br />
Now probability without replacement =$\frac{3\times 2}{4\times 3}=\frac{1}{2}$Probabilityhttp://gateoverflow.in/132635/iisc-phd?show=132672#a132672Sun, 11 Jun 2017 13:32:00 +0000sheldon ross, chapter 2, question 17
http://gateoverflow.in/132359/sheldon-ross-chapter-2-question-17
If 8 rooks are randomly placed on a chessboard, compute the probability that none of the rooks can caputre any of the others. That is compute the probability that no row or file contains more than one rook.Probabilityhttp://gateoverflow.in/132359/sheldon-ross-chapter-2-question-17Fri, 09 Jun 2017 01:32:42 +0000Answered: ACE Practice Test question
http://gateoverflow.in/131760/ace-practice-test-question?show=131812#a131812
<p><strong>P(A/B)=P(A∩B) / P(B)</strong></p>
<p>here given P(B-A)=.15</p>
<p>P(A)=.20</p>
<p>P(A∩B)=.15</p>
<p>as P(B)=P(B-A)+P(A∩B)= .15+.15=.30</p>
<p>so P(A/B)= .15/.30 = .50</p>Probabilityhttp://gateoverflow.in/131760/ace-practice-test-question?show=131812#a131812Sun, 04 Jun 2017 09:04:51 +0000Answered: Probability puzzles
http://gateoverflow.in/130530/probability-puzzles?show=130542#a130542
Answer is 55, solve it using Fibinacci series. F(9)=F(8)+F(7), keep doing it, until F(2)=F(1)+F(0)=1+1=2(as given in the question too that to reach stone 2 we have 2 paths), now trace back F(3) and so on. Final answer is the 9th term in fibonacci series assuming the series begins from 1,2,3,5...55.Probabilityhttp://gateoverflow.in/130530/probability-puzzles?show=130542#a130542Tue, 23 May 2017 12:49:10 +0000Probbility puzzles
http://gateoverflow.in/130529/probbility-puzzles
Three men — conveniently named A, B, and C — are fighting a duel with pistols. It's A's turn to shoot. <br />
<br />
The rules of this duel are rather peculiar: the duelists do not all shoot simultaneously, but instead take turns. A fires at B, B fires at C, and C fires at A; the cycle repeats until there is a single survivor. If you hit your target, you'll fire at the next person on your next turn. <br />
<br />
For example, A might shoot and hit B. With B out of the picture, it would be C's turn to shoot — suppose he misses. Now it's A's turn again, and he fires at C; if he hits, the duel is over, with A the sole survivor. <br />
<br />
To bring in a little probability, suppose A and C each hit their targets with probability 0.5, but that B is a better shot, and hits with probability 0.75 — all shots are independent. <br />
<br />
What's the probability that A wins the duel?Probabilityhttp://gateoverflow.in/130529/probbility-puzzlesTue, 23 May 2017 10:08:48 +0000Answered: Conditional Probability IITB (RA) 2016
http://gateoverflow.in/46150/conditional-probability-iitb-ra-2016?show=130518#a130518
<p><strong>Note:</strong> 1) Brace yourself, spoon feeding is about to begin.</p>
<p> 2) first two dice --- implies two blue dice.</p>
<p> 3rd dice --- implies red dice. </p>
<hr>
<p> </p>
<p><strong>Point 1)</strong> What's asked in question?
<br>
ans) P(person actually won | person claimed that he won) = ?</p>
<p><strong>Point 2)</strong> Why 1/216 isn't the ans.?
<br>
ans) After reading this question, first thought that arrives is....ummm...isn't 1/216 the correct ans. (I bet most people must have thought this).
<br>
But read point 1 (above), point 3 (below) and now read question 2-3 times.
<br>
Makes Sense now?
<br>
(1/216 is just the probability that person won! and this isn't what's asked in the question!)</p>
<p><strong>Point 3) </strong> Elaborating srestha's ans:
<br>
ans)
<br>
<strong>4 cases arise:</strong>
<br>
<br>
person won and 1 came on red dice ---- person tells the truth i.e. person <strong>tells he won</strong>.
<br>
person won and 2 or 3 or 4 or 5 or 6 came on red dice ---- person tells the lie i.e. person <strong>tells he lost</strong>.
<br>
person lost and 1 came on red dice ---- person tells the truth i.e. person <strong>tells he lost</strong>.
<br>
person lost and 2 or 3 or 4 or 5 or 6 came on red dice ---- person tells the lie i.e. person <strong>tells he won</strong>.</p>
<p> </p>
<p> <strong>Therefore,</strong> person claims "he won" in 2 cases:
<br>
<br>
person won and 1 came on red dice
<br>
person lost and 2,3,4,5,6 came on red dice</p>
<p> <strong>Hence, </strong>P(person actually won | person claimed that he won) = ??</p>
<p> P(person actually won and person claimed he won)
<br>
= ---------------------------------------------------------------------------------------------------------------
<br>
P(person actually won and person claimed that he won) + P(person actually lost and person claimed that he won)</p>
<p> (6 came on first dice and 6 came on 2nd dice and 1 came on 3rd dice)
<br>
= -----------------------------------------------------------------------------------------------------------------
<br>
(6 came on first dice and 6 came on 2nd dice and 1 came on 3rd dice) + (double 6 didn't come on first 2 dice throws and (2 or 3 or 4 or 5 or 6 came on 3rd dice))</p>
<p> 1/216
<br>
= ------------------------------
<br>
1/216 + (35/36)(5/36)</p>
<p> 1
<br>
= ------------------------------
<br>
1+175
<br>
<br>
1
<br>
= ------------------------------
<br>
176</p>
<p> </p>
<p><strong>That's it folks!</strong></p>
<p><strong>-----------------------------------------------------------------------------------</strong></p>
<p><strong>and kudos to srestha coming up with the correct ans.</strong></p>
<p><strong>-----------------------------------------------------------------------------------</strong></p>Probabilityhttp://gateoverflow.in/46150/conditional-probability-iitb-ra-2016?show=130518#a130518Tue, 23 May 2017 07:50:09 +0000Rosen Discreate Probability
http://gateoverflow.in/130478/rosen-discreate-probability
Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 99.9% of people with the disease test positive and only 0.02% who do not have the disease test positive.<br />
<br />
a) What is the probability that someone who tests positive has the genetic disease?<br />
<br />
b) What is the probability that someone who tests negative does not have the disease?<br />
<br />
My Answers :-<br />
<br />
P(D) = 0.00001 <br />
<br />
P(D') = 0.99999<br />
<br />
P(P/D) = 0.999<br />
<br />
P(P'/D) = 0.001<br />
<br />
P(P/D') = 0.0002<br />
<br />
P(P'/D') = 0.9998<br />
<br />
Answer a) $\frac{(0.999)(0.00001)}{(0.999)(0.00001) + (0.0002)(0.99999)}$<br />
<br />
Answer b) $\frac{(0.9998)(0.99999)}{(0.9998)(0.99999) + (0.001)(0.00001)}$<br />
<br />
Correct me if I am wrong?Probabilityhttp://gateoverflow.in/130478/rosen-discreate-probabilityMon, 22 May 2017 16:34:47 +0000Answered: counting
http://gateoverflow.in/130233/counting?show=130236#a130236
<table border="1" cellpadding="1" cellspacing="1" style="width:500px">
<tbody>
<tr>
<td> letter I</td>
<td>letter T</td>
<td>letter J</td>
<td>letter E</td>
<td>permutation </td>
<td>result</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>1</td>
<td>4!</td>
<td>24</td>
</tr>
<tr>
<td>2</td>
<td>1</td>
<td>1</td>
<td>0</td>
<td>4!/2!</td>
<td>12</td>
</tr>
<tr>
<td>2</td>
<td>1</td>
<td>0</td>
<td>1</td>
<td>4!/2!</td>
<td>12</td>
</tr>
<tr>
<td>2</td>
<td>0</td>
<td>1</td>
<td>1</td>
<td>4!/2!</td>
<td>12</td>
</tr>
<tr>
<td>0</td>
<td>1</td>
<td>1</td>
<td>2</td>
<td>4!/2!</td>
<td>12</td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td>1</td>
<td>2</td>
<td>4!/2!</td>
<td>12</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>0</td>
<td>2</td>
<td>4!/2!</td>
<td>12</td>
</tr>
<tr>
<td>2</td>
<td>0</td>
<td>0</td>
<td>2</td>
<td>4!/(2! *2!)</td>
<td>6</td>
</tr>
<tr>
<td> </td>
<td> </td>
<td> </td>
<td> </td>
<td>TOTAL ---></td>
<td><span class="marker">102</span></td>
</tr>
</tbody>
</table>
<p><big>102 (ans)</big></p>
<p>c) optionn</p>Probabilityhttp://gateoverflow.in/130233/counting?show=130236#a130236Sat, 20 May 2017 12:24:40 +0000Answered: ISI 2015 MMA 7
http://gateoverflow.in/129269/isi-2015-mma-7?show=129277#a129277
C. $\frac{1-e^{-\lambda}}{\lambda}$<br />
<br />
\begin{align*}<br />
E\left( \frac{1}{1+x}\right ) &= \sum_{k = 0}^{\infty} \left( \frac{1}{1+k}\right ) * \frac{\lambda^k*e^{-\lambda}}{k!} \\<br />
&= \frac{1}{\lambda} * \sum_{k = 0}^{\infty} \frac{\lambda^{k+1}*e^{-\lambda}}{(k+1)!}\\<br />
&= \frac{e^{-\lambda}}{\lambda} * \sum_{k = 1}^{\infty} \frac{\lambda^{k}}{k!}\\<br />
&= \frac{e^{-\lambda}}{\lambda} * \left( \sum_{k = 0}^{\infty} \frac{\lambda^{k}}{k!}-1\right)\\ <br />
&= \frac{e^{-\lambda}}{\lambda} * \left( e^{\lambda} -1\right)\\ <br />
&= \frac{1-e^{-\lambda}}{\lambda}<br />
<br />
\end{align*}Probabilityhttp://gateoverflow.in/129269/isi-2015-mma-7?show=129277#a129277Thu, 11 May 2017 11:35:55 +0000Answered: #probability
http://gateoverflow.in/128939/%23probability?show=129000#a129000
S = 16C8;<br />
A = (2C2x14C6)+(4C4x12C4)+(6C6x10C2) = 3543;<br />
P = A/S = 0.275.<br />
Explanation: Firstly think the probability has to come very less to 1,beacause der r 16 shoes and the pairs r being randomly chosen,so definitely 0.947 is not the answer!LOL!<br />
Now just see der r 16 shoes(8 pairs) ::::<br />
The meaning of 8 shoes randomly selected is,u can choose either 4 pairs or in random way<br />
Now atleast 1 and atmost 3 pairs means, among 16 u can choose 2 shoes(which has to be a pair)<br />
so 2C2,and remaining 6 shoes r randomly selected from 14,so the event for atleast one is::::2C2x14C6<br />
I think this is simple question! and d rest of all is the same way!!!Probabilityhttp://gateoverflow.in/128939/%23probability?show=129000#a129000Tue, 09 May 2017 10:39:09 +0000Answered: IIITH-PGEE 2017
http://gateoverflow.in/127525/iiith-pgee-2017?show=127564#a127564
Conditional Probability : similar question <br />
<br />
<a href="http://gateoverflow.in/41499/gate2014-ec01-ga10?show=41499#q41499" rel="nofollow" target="_blank">http://gateoverflow.in/41499/gate2014-ec01-ga10?show=41499#q41499</a><br />
<br />
Answer : 1/3Probabilityhttp://gateoverflow.in/127525/iiith-pgee-2017?show=127564#a127564Sun, 30 Apr 2017 23:26:34 +0000Answered: GATE2015-1_29
http://gateoverflow.in/8253/gate2015-1_29?show=127559#a127559
The probability of sending a frame in the first slot<br />
without any collision by any of these four stations is<br />
sum of following 4 probabilities<br />
<br />
Probability that S1 sends a frame and no one else does +<br />
Probability thatS2 sends a frame and no one else does +<br />
Probability thatS3 sends a frame and no one else does +<br />
Probability thatS4 sends a frame and no one else does<br />
<br />
= 0.1 * (1 - 0.2) * (1 - 0.3) *(1 - 0.4) +<br />
(1 -0.1) * 0.2 * (1 - 0.3) *(1 - 0.4) +<br />
(1 -0.1) * (1 - 0.2) * 0.3 *(1 - 0.4) +<br />
(1 -0.1) * (1 - 0.2) * (1 - 0.3) * 0.4<br />
<br />
= 0.4404Probabilityhttp://gateoverflow.in/8253/gate2015-1_29?show=127559#a127559Sun, 30 Apr 2017 21:50:14 +0000Answered: GATE2011_18
http://gateoverflow.in/2120/gate2011_18?show=127057#a127057
<p>The difference between (E[X²]) and (E[X])² is called <a rel="nofollow" href="http://en.wikipedia.org/wiki/Variance" target="_blank">variance </a>of a random variable. <strong>V</strong><strong>ariance</strong> measures how far a set of numbers is spread out. (A variance of zero indicates that all the values are identical.) A non-zero variance is always positive:</p>Probabilityhttp://gateoverflow.in/2120/gate2011_18?show=127057#a127057Wed, 26 Apr 2017 22:33:11 +0000Answered: GATE2004-74
http://gateoverflow.in/1068/gate2004-74?show=126620#a126620
Expected marks per question is = -0.25 * 3/4 + 1 * 1/4 = 1/16<br />
Since choice is uniformly distributed, expected marks = 150*1000/16 = 9375Probabilityhttp://gateoverflow.in/1068/gate2004-74?show=126620#a126620Sun, 23 Apr 2017 08:59:55 +0000Answered: GATE2004-25
http://gateoverflow.in/1022/gate2004-25?show=126328#a126328
Another approach is tat in d sample space of Heads and Tails there are 10 placements of Head and Tail which don't contain Two Heads and Two Tails.They are given as follows:<br />
<br />
(T,T,T,T),(H,H,H,H),(T,T,T,H),(T,H,T,T),(T,T,H,T),(H,T,T,T),(H,H,H,T),(H,T,H,H),(H,H,T,H),(T,H,H,H).After having removed these options 6 will be left.Therefore,probability is given by 6/16 which evaluates to 3/8.:)Probabilityhttp://gateoverflow.in/1022/gate2004-25?show=126328#a126328Thu, 20 Apr 2017 21:10:32 +0000Answered: Probability
http://gateoverflow.in/125997/probability?show=126138#a126138
A hybrid parent has rd gene pair and it can contribute either r or d to a child. Considering the scenario where both parents are hybrid i.e rd, both can contribute r or d. The resultant child can have gene pair among these (rr, dd, rd) with rr having probability 1/4, same goes for dd and rd having probability 1/2 as rd and dr would mean the same type of person.<br />
<br />
Now to have an appearance similar to dominance type , child should either be dd or rd whose probability is $\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$ whereas probability of not looking like dominance type is 1/4 (pure recessive rr).<br />
<br />
Probability that 3 of the children have the outward appearance of the dominant gene : $\frac{3}{4} * \frac{3}{4} * \frac{3}{4} * \frac{1}{4}$ * 4 (as any of the 4 children can be of rr type so permutation does not matter)<br />
<br />
The value equals to : $\frac{27}{64}$ This is the required probabilityProbabilityhttp://gateoverflow.in/125997/probability?show=126138#a126138Wed, 19 Apr 2017 00:02:29 +0000Answered: Probability
http://gateoverflow.in/125714/probability?show=126120#a126120
<p>Probability of one dealt is $\frac{Total possible full house pairs}{Total possible pairs}$</p>
<p>Total Possible full house pair = 13*12*24=3744</p>
<p>For 3 of an kind we have to choose 2 different denomination of card i.e (3,4)= 3,3,3,4,4 but (4,3)=4,4,4,3,3 thus order matters so this can be done by <sup>13</sup>P<sub>2</sub> ways =13*12</p>
<p>form 4 cards of same denomination we can choose 3 in <sup>4</sup>C<sub>3</sub> ways =4</p>
<p>form 4 cards of same denomination we can choose 2 in <sup>4</sup>C<sub>2</sub> ways =6</p>
<p>Total possible full house is 13*12*4*6=3744</p>
<p>Finding total possible combination is easy we have 52 cards and 5 positions thus<sup>52</sup>C<sub>5</sub> ways</p>
<p>P(Full House)=$\frac{3744}{^{52}C_{5}}$</p>
<p> </p>Probabilityhttp://gateoverflow.in/125714/probability?show=126120#a126120Tue, 18 Apr 2017 19:52:21 +0000Answered: Probability
http://gateoverflow.in/125717/probability?show=125783#a125783
<p>The soln in my terms </p>
<p>1) let there are 4 players A,B,C,D </p>
<p>For player A </p>
<p>First card a spade 13/52 second card also spade =12/51 and so on .... so 13x12x11x..1/(52x51x50...40)</p>
<p>and similarly for rest B,C,D so ans is 4 x(13 x12 x11....1) /(52x51x...40)</p>
<p>2)For player A having exactly one ace = 12 non ace and one ace at any location 12 non-ace=48x47x46x...37/52x51....41 one ace =4/40 =1/10 now ace can take any 13 position so </p>
<p>prob of A having exactly one ace =13 (48 x47x 46 x....37x4) /(52x51x50....41x40)</p>
<p>and same for B , C ,D so </p>
<p>ans= {13 (48 x47x 46 x....37x4) /(52x51x50....41x40)}<sup>4</sup></p>
<p> </p>
<p>for more details refer <a rel="nofollow" href="https://math.la.asu.edu/~quigg/teach/courses/421/2014fall/notes/stp421notes.pdf">https://math.la.asu.edu/~quigg/teach/courses/421/2014fall/notes/stp421notes.pdf</a></p>Probabilityhttp://gateoverflow.in/125717/probability?show=125783#a125783Sun, 16 Apr 2017 10:52:08 +0000Answered: GATE2000-2.2
http://gateoverflow.in/649/gate2000-2-2?show=125776#a125776
<p>Given Constraints:</p>
<p>1. Pr(E1) = Pr(E2)</p>
<p>2. Pr( E1 U E2) = 1</p>
<p>3. E1 and E2 are independent</p>
<p>As we know: Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1 ∩ E2) As E1 and E2 are independent events. (cond.3)</p>
<p>So Pr(E1 ∩ E2) = Pr(E1) Pr(E2) Pr(E1) = Pr(E2) (cond.2)</p>
<p>let probability of Event E1 = x = prob of E2 So, Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1) Pr(E2) 1 = x + x -x* x (cond. 1) 1=2x-x^2 x^2-2x+1 = 0 (x-1)^2 = 0 x = 1 So, Pr(E1) = Pr(E2) = 1 Thus, option (D) is the answer.</p>
<p>Reference : <a rel="nofollow" href="https://people.richland.edu/james/lecture/m170/ch05-rul.html">https://people.richland.edu/james/lecture/m170/ch05-rul.html</a></p>
<p><strong>Another Solution :</strong> E1 and E2 are independent events.
<br>
Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1) Pr(E2)
<br>
Pr(E1) = Pr(E2) (given)
<br>
So,
<br>
2 * Pr(E1) – Pr(E1)<sup>2</sup> = Pr( E1 U E2)
<br>
2 * Pr(E1) – Pr(E1)<sup>2</sup> = 1
<br>
So, Pr(E1) = Pr(E2) = 1
<br>
Thus, option (D) is the answer.</p>Probabilityhttp://gateoverflow.in/649/gate2000-2-2?show=125776#a125776Sun, 16 Apr 2017 10:02:59 +0000Answered: GATE2007-24
http://gateoverflow.in/1222/gate2007-24?show=125473#a125473
<p>Here order of odd numbers doesn't matter, so we focus only on 10 even numbers as if we have to arrange only 10 even numbers . In general, digit 2 can be placed at any of the 10 places available, but according to question, 2 can be placed at only 1st because it has to appear before every other even number. So out of 10 choices, we have only 1 favourable choice, so probability is 1/10. So option <strong>(B)</strong> is correct.</p>Probabilityhttp://gateoverflow.in/1222/gate2007-24?show=125473#a125473Fri, 14 Apr 2017 09:24:31 +0000Answered: Probability Exercise
http://gateoverflow.in/123094/probability-exercise?show=125174#a125174
<p>We have two sets : one is having 8 elements and another one is having 9 elements. (sets representing the schools here)</p>
<p>We are interested in two particular element x and y (for example in place of Rebecca and Elise) </p>
<p>These two elements $(x,y)$ belong to different sets. Allocate them in any order in those two sets (schools are indistinct, therefore we need not swap $x$,$y$ and simulate the following process again)</p>
<p> </p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=2128306449176225140"></p>
<p> </p>
<p>After selecting $4$ elements from each set we can map $(x,a,b,c)$ to $(y,p,q,r)$ in $4!$ ways in a one-to-one and onto mapping. These $4!$ ways also include $3!$ ways in which $x$ is always mapped to $y$ and $3*3!$ ways in which $x$ is never mapped to $y$.</p>
<p>(a).</p>
<p>probability that (x,y) pair results at the end = </p>
<p>$\begin{align*}
<br>
\left [ \text{probability of x getting selected} \right ] \cdot \left [ \text{probability of y getting selected} \right ] \cdot \left [ \text{probability of getting (x,y) pair} \right ]
<br>
\end{align*}$</p>
<p>=</p>
<p>$\begin{align*}
<br>
\left [ \frac{7C3}{8C4} \right ] \cdot \left [ \frac{8C3}{9C4} \right ] \cdot \left [ \frac{3!}{4!} \right ] = \frac{1}{18} \\
<br>
\end{align*}$</p>
<p> </p>
<hr>
<p> </p>
<p>(b).</p>
<p>probability that (x,y) pair does not result at the end =</p>
<p>$\begin{align*}
<br>
\left [ \frac{7C3}{8C4} \right ] \cdot \left [ \frac{8C3}{9C4} \right ] \cdot \left [ \frac{3*3!}{4!} \right ] = \frac{1}{6} \\
<br>
\end{align*}$</p>
<p> </p>Probabilityhttp://gateoverflow.in/123094/probability-exercise?show=125174#a125174Tue, 11 Apr 2017 23:42:35 +0000Answered: TIFR2011-A-19
http://gateoverflow.in/26479/tifr2011-a-19?show=125167#a125167
Case 1:largest is 5 , smallest 1 and middle is 2 or 3 or 4 : 3*3!<br />
<br />
Case 2.largest is 5 , smallest 1 and middle is 1 or 5 : 3!*2/2!<br />
<br />
Case 3:largest is 6 , smallest 2 and middle is 3 or 4 or 5 : 3*3!<br />
<br />
Case 4:largest is 6 , smallest 2 and middle is 6 or 2: 3!*2/2!<br />
<br />
So probability the highest and the lowest value differ by 4 =( 3*3!+3!*2/2!+3*3!+3!*2/2!)/6^3 =2/9Probabilityhttp://gateoverflow.in/26479/tifr2011-a-19?show=125167#a125167Tue, 11 Apr 2017 21:52:58 +0000Answered: GATE2014-1-48
http://gateoverflow.in/1927/gate2014-1-48?show=124568#a124568
In general, Probability (of an event ) = No of favorable outcomes to the event / Total number of possible outcomes in the random experiment.<br />
<br />
Here, 4 six-faces dices are tossed, for one dice there can be 6 equally likely and mutually exclusive outcomes. T<br />
<br />
aking 4 together, there can be total number of 6*6*6*6 = 1296 possible outcomes.<br />
<br />
Now, No of favorable cases to the event : here event is getting sum as 22. So, there can be only 2 cases possible.<br />
<br />
Case 1: Three 6's and one 4, for example: 6,6,6,4 ( sum is 22) Hence, No of ways we can obtain this = 4!/3! = 4 ways ( 3! is for removing those cases where all three 6 are swapping among themselves)<br />
<br />
Case 2: Two 6's and two 5's,for example: 6,6,5,5 ( sum is 22) Hence, No of ways we can obtain this = 4! /( 2! * 2!) = 6 ways ( 2! is for removing those cases where both 6 are swapping between themselves, similarly for both 5 also)<br />
<br />
Hence total no of favorable cases = 4 + 6 = 10. Hence probability = 10/1296. Therefore option D.Probabilityhttp://gateoverflow.in/1927/gate2014-1-48?show=124568#a124568Fri, 07 Apr 2017 08:58:22 +0000Answered: GATE2014-1-2
http://gateoverflow.in/1717/gate2014-1-2?show=124464#a124464
Answer is A<br />
<br />
Source - math.stackexchange.com/questions/350679/we-break-a-unit-length-rod-into-two-pieces-at-a-uniformly-chosen-point-find-the/350863#350863Probabilityhttp://gateoverflow.in/1717/gate2014-1-2?show=124464#a124464Thu, 06 Apr 2017 17:16:34 +0000Answered: ISRO2014-37
http://gateoverflow.in/15032/isro2014-37?show=124279#a124279
Answer is BProbabilityhttp://gateoverflow.in/15032/isro2014-37?show=124279#a124279Wed, 05 Apr 2017 19:13:52 +0000Answered: GATE2017-2-26
http://gateoverflow.in/118368/gate2017-2-26?show=123985#a123985
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=5877261905515026736"></p>
<p>hope it might help........</p>Probabilityhttp://gateoverflow.in/118368/gate2017-2-26?show=123985#a123985Tue, 04 Apr 2017 16:06:31 +0000Answered: GATE2014-2-48
http://gateoverflow.in/2014/gate2014-2-48?show=123949#a123949
<pre>
There are total 100 numbers, out of which
50 numbers are divisible by 2,
33 numbers are divisible by 3,
20 numbers are divisible by 5
Following are counted twice above
16 numbers are divisible by both 2 and 3
10 numbers are divisible by both 2 and 5
6 numbers are divisible by both 3 and 5
Following is counted thrice above
3 numbers are divisible by all 2, 3 and 5
</pre>
<p>So total numbers divisible by 2, 3 and 5 are = = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 103 - 29 = 74 So probability that a number is number is not divisible by 2, 3 and 5 = (100 - 74)/100 = 0.26</p>Probabilityhttp://gateoverflow.in/2014/gate2014-2-48?show=123949#a123949Tue, 04 Apr 2017 13:23:09 +0000Explanation needed!!
http://gateoverflow.in/123899/explanation-needed
<p>You roll a fair four-sided die. If the result is 1 or 2, you roll once
<br>
more but otherwise, you stop. What is the probability that the sum total of your
<br>
rolls is at least 4?</p>
<ol>
<li>9/8</li>
<li>9/16</li>
<li>3/5</li>
<li>1/16</li>
</ol>
<p> </p>Probabilityhttp://gateoverflow.in/123899/explanation-neededTue, 04 Apr 2017 10:39:42 +0000Answered: GATE2014-2-2
http://gateoverflow.in/1954/gate2014-2-2?show=123717#a123717
There are total 9 words in the sentence. To find expected length first we count total length of the sentence i.e 35. Now to find expected length divide total length by total words i.e 35/9 so answer is 3.8 or 3.9 Another Explanation: Expected value = ∑( x * P(x) ) = 3*4/9 + 4*2/9 + 5*3/9 = 35/9 = 3.9Probabilityhttp://gateoverflow.in/1954/gate2014-2-2?show=123717#a123717Mon, 03 Apr 2017 14:16:58 +0000Answered: GATE-2014-2-1
http://gateoverflow.in/1953/gate-2014-2-1?show=123716#a123716
<pre>
Total ways to pick 4 computers = 10*9*8*7
Total ways that at least three computers are fine =
Total ways that all 4 are fine + Total ways any 3 are fine
Total ways that all 4 are fine = 4*3*2*1
Total ways three are fine = 1st is Not working and other 3 working +
2nd is Not working and other 3 working +
3rd is Not working and other 3 working +
4th is Not working and other 3 working +
= 6*4*3*2 + 4*6*3*2 + 4*3*6*2 + 4*3*2*6
= 6*4*3*2*4
The probability = Total ways that at least three computers are fine /
Total ways to pick 4 computers
= (4*3*2*1 + 6*4*3*2*4) / (10*9*8*7)
= (4*3*2*25) / (10*9*8*7)
= 11.9% </pre>Probabilityhttp://gateoverflow.in/1953/gate-2014-2-1?show=123716#a123716Mon, 03 Apr 2017 14:10:26 +0000Answered: GATE2014-3-48
http://gateoverflow.in/2082/gate2014-3-48?show=123234#a123234
Sample Space(S) - A set of all possible outcomes/events of a random experiment. Mutually Exclusive Events - Those events which can't occur simultaneously. P(A)+P(B)+P(A∩B)=1 Since the events are mutually exclusive, P(A∩B)=0. Therefore, P(A)+P(B)=1 Now, we now that AM >= GM So, (P(A)+P(B))/2 >= sqrt(P(A)*P(B)) P(A)*P(B) <= 1/4<br />
Hence max(P(A)*P(B)) = 1/4. <br />
We can think of this problem as flipping a coin, it has two mutually exclusive events ( head and tail , as both can't occur simultaneously). And sample space S = { head, tail } Now, lets say event A and B are getting a "head" and "tail" respectively. Hence, S = A U B. Therefore, P(A) = 1/2 and P(B) = 1/2. And, P(A).P(B) = 1 /4 = 0.25. Hence option B is the correct choice.Probabilityhttp://gateoverflow.in/2082/gate2014-3-48?show=123234#a123234Sat, 01 Apr 2017 11:28:59 +0000Answered: Ravi asked his neighbor to water a delicate plant while he is away.
http://gateoverflow.in/122637/ravi-asked-his-neighbor-to-water-delicate-plant-while-he-away?show=122666#a122666
<p>Draw the diagram:</p>
<p><img alt="" src="http://gateoverflow.in/?qa=blob&qa_blobid=6963821512883465008"></p>
<p> </p>
<p>Required probability = $$\begin{align*} \frac{0.1 \cdot 0.8}{0.1 \cdot 0.8 + 0.9 \cdot 0.15} = 0.372 \end{align*}$$</p>
<p> </p>
<p>Answer = $16/43$</p>
<p> </p>Probabilityhttp://gateoverflow.in/122637/ravi-asked-his-neighbor-to-water-delicate-plant-while-he-away?show=122666#a122666Mon, 27 Mar 2017 11:58:30 +0000Answered: The probability of solving a problem by three students X,Y,Z are 1/4,3/7and 2/9 respectively. if all of them try independently then the probability that the problem could not be solved is a)1/3 b)1/4 c)1/5 d) 2/5 e) 3/7
http://gateoverflow.in/5128/probability-students-respectively-independently-probability?show=122105#a122105
1/3 is the correct AnswerProbabilityhttp://gateoverflow.in/5128/probability-students-respectively-independently-probability?show=122105#a122105Sun, 19 Mar 2017 20:35:11 +0000Answered: The probability that A hits a target is 1 / 4 and the probability that B hits a target is 1/3
http://gateoverflow.in/121854/probability-that-hits-target-probability-that-hits-target?show=121879#a121879
$Prob(\text{A hits target = A}) = \frac{1}{4}$<br />
<br />
$Prob(\text{B hits target = B}) = \frac{1}{3}$<br />
<br />
Given ,that Target is hit only once:<br />
<br />
$ Prob(\text{Target is hit only once = O}) = \frac{1}{4}*\frac{2}{3}+\frac{3}{4}*\frac{1}{3} = \frac{10}{24} $<br />
<br />
Probability, that A hits the target, given that the target was hit only once:<br />
<br />
$Prob(A|O) = \frac{Prob(A \cap O)}{Prob(O)} = \frac{1}{6}*\frac{24}{10} = \frac{4}{10} = 0.4 $Probabilityhttp://gateoverflow.in/121854/probability-that-hits-target-probability-that-hits-target?show=121879#a121879Fri, 17 Mar 2017 08:31:06 +0000probability
http://gateoverflow.in/121747/probability
If from each box of the three boxes containing $3$ white and $3$ black,$2$ white and $2$ black,$1$ white and $3$ black,one ball is drawn at random ,then the probability that $2$ white and $1$ black ball will be drawn ?Probabilityhttp://gateoverflow.in/121747/probabilityThu, 16 Mar 2017 11:35:10 +0000