GATE Overflow - Recent questions and answers in Probability
https://gateoverflow.in/qa/mathematics/probability
Powered by Question2AnswerAnswered: test series
https://gateoverflow.in/172351/test-series?show=173178#a173178
Answer is 0.96?Probabilityhttps://gateoverflow.in/172351/test-series?show=173178#a173178Wed, 22 Nov 2017 18:20:03 +0000Probability
https://gateoverflow.in/171698/probability
In an area, it rains once in every three days. The local evening newspaper attempts to predict whether it rains or not in the following day. Three quarters of rainy days and three fifths of dry days are correctly predicted by the previous evening’s paper. Given that this evening’s paper predicts rain, what is the probability that it will actually rain tomorrow?Probabilityhttps://gateoverflow.in/171698/probabilitySat, 18 Nov 2017 17:34:20 +0000Answered: GATE2015-3_37
https://gateoverflow.in/8496/gate2015-3_37?show=171621#a171621
<p>We have to find $P\left ( \frac{Y=0}{X_{3}=0} \right ) = ? $ </p>
<p> </p>
<table border="1" cellpadding="1" cellspacing="1" style="width:500px">
<tbody>
<tr>
<td>$X_{1}$</td>
<td>$X_{2}$</td>
<td>$X_{3}$</td>
<td>$Y$</td>
</tr>
<tr>
<td>0</td>
<td>0</td>
<td><span class="marker">0</span></td>
<td><span class="marker">0</span></td>
</tr>
<tr>
<td>0</td>
<td>0</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>0</td>
<td>1</td>
<td><span class="marker">0</span></td>
<td><span class="marker">0</span></td>
</tr>
<tr>
<td>0</td>
<td>1</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td><span class="marker">0</span></td>
<td><span class="marker">0</span></td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td><span class="marker">0</span></td>
<td>1</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>0</td>
</tr>
</tbody>
</table>
<p> </p>
<p>$P\left ( Y=0 \cap X_{3}=0\right ) = \frac{3}{8}$ and $P\left ( X_{3} = 0 \right ) = \frac{4}{8}$</p>
<p> </p>
<p>So $P\left ( \frac{Y=0}{X_{3}=0} \right ) = \frac{P\left ( Y=0 \cap X_{3}=0\right )}{P\left ( X_{3} = 0 \right )} = \frac{3}{4}$</p>
<p> </p>
<p> </p>Probabilityhttps://gateoverflow.in/8496/gate2015-3_37?show=171621#a171621Sat, 18 Nov 2017 14:46:56 +0000Answered: TIFR2012-A-1
https://gateoverflow.in/20938/tifr2012-a-1?show=170679#a170679
<p>Tree-Diagram in such problems make them easier to solve</p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=14186573561546279946"></p>
<p><strong>Let's say India Won.</strong></p>
<p>In Two ways probability can be assigned to India's Win</p>
<p> </p>
<p>The two favorable cases of our problem are encircled by green.</p>
<p><strong>Case 1: Both Amar and Akbar Tell Truth</strong></p>
<p>Actual Case: India Won</p>
<p>Amar Tell Akbar: India Won</p>
<p>Akbar Tells Anthony : Amar Told me India Won <strong>(Yes this is true)</strong></p>
<p>So probability for this = $\frac{3}{4} * \frac{3}{4} = \frac{9}{16}$</p>
<p><strong>Case 2: Amar Tells lies and Akbar also Lies</strong></p>
<p>Actual Case: India Won</p>
<p>Amar Told Akbar: India Lost</p>
<p>Akbar Told Anthony: Amar told me India Won </p>
<p>So, Reversing statement of Akbar we get "Amar told me India lost"</p>
<p>This means Amar Told "India has lost" to Akbar</p>
<p>Since Amar also lied, so This means India Must have Won!!.</p>
<p>So, Probability for this case = $\frac{1}{4} * \frac{1}{4} = \frac{1}{16}$</p>
<p>Total Probability = $\frac{10}{16}$</p>
<p> </p>Probabilityhttps://gateoverflow.in/20938/tifr2012-a-1?show=170679#a170679Thu, 16 Nov 2017 05:56:38 +0000Answered: CMI2013-A-02
https://gateoverflow.in/46592/cmi2013-a-02?show=170671#a170671
<p>Such questions are much easier to solve once, you build a tree and solve it using basic definition of conditional probability.</p>
<p>It is asked that</p>
<p>P(E-mail is reall Spam | Spam Marked by filter) which is given by</p>
<p>$\frac{P(Really Spam \cap Spam marked by filter )}{P(Spam marked by filter)}$</p>
<p>Now look at the below expression tree</p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=13533385261782538512"></p>
<p>The Numerator is the first red box</p>
<p>And denominator term is the sum of two red boxes</p>
<p>P(Really Spam $\cap$ Spam Marked By Filter) = 0.9*0.1=0.09</p>
<p>P(Spam Marked by filter)= P(E-mail was not spam but marked as Spam by filter) + P(E-mail was Spam and marked as Spam by Filter)</p>
<p> = (0.9*0.1)+(0.1*0.9)=0.18</p>
<p>So therefore </p>
<p>$\frac{P(Really Spam \cap Spam marked by filter )}{P(Spam marked by filter)}$ = $\frac{0.09}{0.18} * 100(For Percent)$ = <strong>50%</strong></p>Probabilityhttps://gateoverflow.in/46592/cmi2013-a-02?show=170671#a170671Thu, 16 Nov 2017 05:23:20 +0000Gate previous year
https://gateoverflow.in/169925/gate-previous-year
<p> </p>
<p> GATE2008 QUESTION......ans is given c option but the value of probability that we get on putting random valus of r is -ve......then how can c option be the answer........GUYS PLZZ HELP......<img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=1255963995518535773">b</p>Probabilityhttps://gateoverflow.in/169925/gate-previous-yearTue, 14 Nov 2017 05:40:53 +0000#PROBABILITY
https://gateoverflow.in/169628/%23probability
Is there any relation between independent events and mutually exclusive events?Probabilityhttps://gateoverflow.in/169628/%23probabilityMon, 13 Nov 2017 10:08:47 +0000Answered: GATE2015-1_29
https://gateoverflow.in/8253/gate2015-1_29?show=169484#a169484
<p>Pi is the probability of frame being generated by station si</p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=406831075145364071"></p>Probabilityhttps://gateoverflow.in/8253/gate2015-1_29?show=169484#a169484Mon, 13 Nov 2017 04:01:03 +0000Normal Distribution
https://gateoverflow.in/169307/normal-distribution
Can any one say how to solve Z score questions like below<br />
<br />
1) P(1≤Z≤30) = ?<br />
<br />
2) P(z>2.5)=?<br />
<br />
3) P(z<=k) = 0.9 find k?<br />
<br />
Will they give any related values of Z-scores in exam ?Probabilityhttps://gateoverflow.in/169307/normal-distributionSun, 12 Nov 2017 14:11:09 +0000Answered: probability
https://gateoverflow.in/168943/probability?show=168955#a168955
<p>P(heads) = P</p>
<p>P(tails) = 1-P</p>
<p>Coin is tossed even no of times means 2,4,6,8 ....... </p>
<p>For any throw first head should appear after that tails on rest of the throws in individual trial.</p>
<p>P(E) = P(H)P(T) + P(H)P(T)P(T)P(T) + P(H)P(T)P(T)P(T)P(T)P(T)......</p>
<p>P(E)=$P(1-P) + P(1-P)^{3} + P(1-P)^{5} + ........ ( infinite\ GP)$</p>
<p>P(E) = $\frac{P(1-P)}{1-(1-P)^{2}} = \frac{P(1-P)}{(1-1+P)(1+1-P)} = \frac{1-P}{2-P}$</p>
<p><strong>Hence option b)</strong> <strong>is</strong> <strong>correct</strong></p>Probabilityhttps://gateoverflow.in/168943/probability?show=168955#a168955Sat, 11 Nov 2017 18:11:57 +0000Finding minima of function.
https://gateoverflow.in/168862/finding-minima-of-function
<p>For what value of 'k' (in terms of $X_{i}$'s) following functions will attain their minimum value $\rightarrow$ </p>
<ol>
<li>$F_{1}(k)$ = $\sum_{i=0}^{n} \left | X_{i} - k \right |$</li>
<li>$F_{2}(k)$ = $\sum_{i=0}^{n}\left ( X_{i} - k \right )^{2}$</li>
</ol>
<p>
<br>
If possible, please provide some valid proof for supporting your answer ?</p>Probabilityhttps://gateoverflow.in/168862/finding-minima-of-functionSat, 11 Nov 2017 15:02:05 +0000Range of covariance and correlation coefficient
https://gateoverflow.in/168800/range-of-covariance-and-correlation-coefficient
Hi Guys,<br />
<br />
How can we prove following $\rightarrow$<br />
<br />
$-\infty \leq Covariance(X,Y) \leq +\infty$<br />
<br />
and <br />
<br />
$ -1 \leq correlationCoefficient(X,Y) \leq +1$Probabilityhttps://gateoverflow.in/168800/range-of-covariance-and-correlation-coefficientSat, 11 Nov 2017 12:31:11 +0000Answered: A First Course in Probability- Sheldon Ross 8th Edition Random Variables Example 1b
https://gateoverflow.in/168323/course-probability-sheldon-edition-random-variables-example?show=168356#a168356
<p>This question is from hypergeometric distribution basically which has the following features :</p>
<blockquote>
<p>a) The populatin size which is also known sample space size is finite .</p>
<p>b) Objects are drawn without replacement i.e. once an object is drawn is not put back..</p>
<p>c) Hence probability of drawing each object varies unlike binomial distribution..</p>
</blockquote>
<p>So here we have :</p>
<p>P(at least one ball value >= 17) = 1 - P(all ball values <= 16)</p>
<p> = 1 - ( <sup>16</sup>C<sub>3</sub> / <sup>20</sup>C<sub>3</sub> )</p>
<p> = 1 - [ ( 16 * 15 * 14 ) / (20 * 19 * 18) ]</p>
<p> = 0.50877</p>
<p> = 0.509</p>
<p><em><strong>Hence required probability = 0.509</strong></em></p>Probabilityhttps://gateoverflow.in/168323/course-probability-sheldon-edition-random-variables-example?show=168356#a168356Fri, 10 Nov 2017 11:15:32 +0000Answered: GATE2005-52
https://gateoverflow.in/1177/gate2005-52?show=168256#a168256
let us suppose if outcome is head =>0, tail => 1 Since the coin is fare, P(H) = P(T) = 1⁄2 Length of the string is => n P(X) = both the strings should not be identical P(-X) = both are not identical = 1 – P(X) If both the strings are equal, every character should be same w.r.t its positions i.e P(X) = 1/2*1/2*.......(n times) = (1/2)^n P(-X) = 1 – (1/2)^nProbabilityhttps://gateoverflow.in/1177/gate2005-52?show=168256#a168256Fri, 10 Nov 2017 07:05:03 +0000Answered: Probability Distribution
https://gateoverflow.in/166925/probability-distribution?show=166985#a166985
<p><a rel="nofollow" href="https://www.khanacademy.org/math/statistics-probability/random-variables-stats-library/poisson-distribution/v/poisson-process-1">https://www.khanacademy.org/math/statistics-probability/random-variables-stats-library/poisson-distribution/v/poisson-process-1</a></p>Probabilityhttps://gateoverflow.in/166925/probability-distribution?show=166985#a166985Mon, 06 Nov 2017 18:35:53 +0000Answered: PROBABILITY
https://gateoverflow.in/166680/probability?show=166864#a166864
<p>'A' and 'B' throws the dice alternatively and the sequence is ABABAB................</p>
<p>'A' wins sum of the dice is 8.</p>
<p>and probability of getting 8 is P(A)=5/36. [(2,6),(3,5),(4,4),(5,3),(6,2)]</p>
<p>same as 'B' wins sum of two dice is 7 and probability of getting sum 7 is P(B)=6/36.</p>
<p><strong>Probability of 'A' wins the game is.</strong></p>
<p>=P(A)+P($\bar{A}$)P($\bar{B}$)P(A)+P($\bar{A}$)P($\bar{B}$)P($\bar{A}$)P($\bar{B}$)P(A)+.........................</p>
<p>=$\frac{5}{36}$+$\frac{31}{36}*\frac{30}{36}*\frac{5}{6}$+$\frac{31}{36}*\frac{30}{36}*\frac{31}{36}*\frac{30}{36}*\frac{5}{6}$+.........</p>
<p>=$\frac{5}{6}$[1+$\frac{31}{36}*\frac{30}{36}$+$\frac{31}{36}*\frac{30}{36}*\frac{31}{36}*\frac{30}{36}$+.................]</p>
<p>=$\frac{5}{36}*(1-(\frac{31*30}{36*36}))^{-1}$</p>
<p>=$\frac{5}{36}*(\frac{36*36}{36*36-31*30})$</p>
<p>=0.49180</p>
<p>=$\frac{30}{61}$.</p>
<p><strong>Hence option should be C</strong></p>Probabilityhttps://gateoverflow.in/166680/probability?show=166864#a166864Mon, 06 Nov 2017 13:17:31 +0000Answered: TIFR2010-B-38
https://gateoverflow.in/19050/tifr2010-b-38?show=166625#a166625
<p>We have coin lying on table.</p>
<p>we have to select one coin ad then toss it.</p>
<p>In starting <strong>we have {H,H,T } outcome</strong></p>
<p>Now from here we can only select one coin then flip it.</p>
<p>So,number of all posible outcomes are those which have<strong> 1 or 0 hamming distance from above outcome.</strong></p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=5297440175799531746"></p>
<p>Here total outcomes are ---6</p>
<p>And outcomes with at least two heads are----4</p>
<p><strong>Ans is=4/6=2/3</strong></p>Probabilityhttps://gateoverflow.in/19050/tifr2010-b-38?show=166625#a166625Mon, 06 Nov 2017 04:10:17 +0000Answered: GATE2011_33
https://gateoverflow.in/2135/gate2011_33?show=165940#a165940
<p>(D) Answer. σy=aσx</p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=7209320049910249410"></p>Probabilityhttps://gateoverflow.in/2135/gate2011_33?show=165940#a165940Sat, 04 Nov 2017 14:23:39 +0000Answered: GATE1995_2.14
https://gateoverflow.in/2626/gate1995_2-14?show=165819#a165819
Probability that both balls are black= (15C2)/(25C2) =7/20<br />
<br />
Probability that both balls are white= (10C2)/(25C2) =6/40<br />
<br />
Probability that (both balls are black) or (both balls are white)= 7/20 + 6/40 =20/40=1/2<br />
<br />
The probability that one of them is black and the other is white is: 1- 1/2 = 1/2Probabilityhttps://gateoverflow.in/2626/gate1995_2-14?show=165819#a165819Sat, 04 Nov 2017 06:22:00 +0000Answered: GATE2011_3
https://gateoverflow.in/2105/gate2011_3?show=165769#a165769
When two coins are flipped , sample space is { HH, HT, TH, TT }<br />
<br />
Now if it is known that at least one of them is head then, reduced sample space will be --> { HH, HT, TH}<br />
<br />
In this reduced sample space only one sample point fulfilling the criteria( both are heads HH) hence required probability is 1/3Probabilityhttps://gateoverflow.in/2105/gate2011_3?show=165769#a165769Sat, 04 Nov 2017 04:04:09 +0000Answered: GATE2014-3-48
https://gateoverflow.in/2082/gate2014-3-48?show=165767#a165767
When P(A)P(B) is maximum A,B together should complete whole universal space( P(A∪B)=1) and P(A),P(B) not equal to 0<br />
<br />
It is given that A,B are mutually exclusive events(P(A∪B)= P(A) + P(B) ). So, if we let P(A)=x then, <br />
<br />
P(A∪B)= P(A) + P(B) or, 1= x +P(B) or, P(B)=1-x. Hence, P(A)P(B)= x.(1-x)<br />
<br />
derivative of x(1-x)= 1-2x, So, when P(A)P(B) is maximum, 1-2x=0 or, x=1/2, Hence x(1-x)= 1/2 * 1/2 = 1/4=0.25Probabilityhttps://gateoverflow.in/2082/gate2014-3-48?show=165767#a165767Sat, 04 Nov 2017 03:54:57 +0000Answered: TIFR2015-A-1
https://gateoverflow.in/29156/tifr2015-a-1?show=165431#a165431
P(2,4,6) =1/2 , P(3,6)= 1/3 , P(1)= 1/6 <br />
<br />
So, P(2,4,6) +P(3,6) - P(6) + P(1) + P(5) =1<br />
<br />
or, 1/2 + 1/3 - P(6) + 1/6 + P(5) =1<br />
<br />
or, 1 - P(6) + P(5) =1<br />
<br />
or, P(5) =P(6)<br />
<br />
Now we know, P(3,6)=1/3<br />
<br />
or, P(3) + P(6) = 1/3<br />
<br />
or, P(3) + P(5) =1/3 [ since P(5)=P(6)]<br />
<br />
or, P(5) =1/3 - P(3)<br />
<br />
Since P(3)>=0, P(5)<=1/3Probabilityhttps://gateoverflow.in/29156/tifr2015-a-1?show=165431#a165431Fri, 03 Nov 2017 05:14:57 +0000Answered: GATE2004-78
https://gateoverflow.in/1072/gate2004-78?show=165321#a165321
Number of binary string pairs (S1,S2) each having length n ,can be formed in 2^n * 2^n ways.<br />
<br />
Now we have to find out the number of such pairs where hamming distance between S1,S2 is d.<br />
<br />
In order to form such pairs first select any S1(which can be done is 2^n ways),<br />
<br />
Now lets form S2: Select nCd positions from S2, once nCd bit positions from S2 is selected, for those positions we have only one choice(opposite of S1's value for that particular position). Remaining n-d places of S2 will have exactly same value as S1's value for those positions(this will make sure that hamming distance of S1, S2 is exactly d) hence only one way we can do so.<br />
So , for a particular S1, we can form S2 in nCd *1 *1=nCd ways ,such that hamming distance of (S1,S2)=d.<br />
<br />
There are 2^n ways to form S1, So total number of pairs where hamming distance is d = 2^n * nCd<br />
<br />
We already know number of pairs each having length n ,can be formed in 2^n * 2^n ways.<br />
<br />
Hence required probability= (2^n * nCd) / ( 2^n * 2^n) = nCd/ 2^nProbabilityhttps://gateoverflow.in/1072/gate2004-78?show=165321#a165321Thu, 02 Nov 2017 17:32:21 +0000Answered: Gate 2012 CE Probability
https://gateoverflow.in/164724/gate-2012-ce-probability?show=164729#a164729
<p>Given data</p>
<p>$\mu = 1000 mm$, $\sigma = 200 mm$</p>
<p>Converting into normal form,</p>
<p>Z = $\frac{X-\mu }{\sigma }$ = $\frac{X-1000}{200}$</p>
<p>Now at X = 1200, Z = $\frac{1200-1000}{200}$ = 1</p>
<p>we want P(X > 1200) = P(Z > 1) which is definitely less than 50% as we know for standard normal variate Z, P(Z > 0) = 50 %</p>
<p><strong>A is the answer</strong></p>Probabilityhttps://gateoverflow.in/164724/gate-2012-ce-probability?show=164729#a164729Wed, 01 Nov 2017 09:16:26 +0000Answered: Probability
https://gateoverflow.in/164529/probability?show=164543#a164543
sum 6<br />
<br />
1,5<br />
<br />
2,4<br />
<br />
3,3<br />
<br />
4,2<br />
<br />
5,1<br />
<br />
probability of getting 4 on first dice = 1/5Probabilityhttps://gateoverflow.in/164529/probability?show=164543#a164543Tue, 31 Oct 2017 16:43:22 +0000Answered: GATE2012_33
https://gateoverflow.in/1751/gate2012_33?show=164088#a164088
Let, Ei be the event that i shows(1<= i <= 6) when a die is rolled. Clearly ,P(Ei)=1/6<br />
<br />
Let, A be the event that sum is at least 6.<br />
<br />
we need to find , P(E6)+ P(E1 ∩ A) + P(E2 ∩ A) + P(E3 ∩ A)<br />
<br />
= 1/6 + P(E1) P(A/E1) + P(E2) P(A/E2) + P(E3) P(A/E3)<br />
<br />
=1/6 + 1/6 * 2/6 + 1/6 * 3/6 + 1/6 * 4/6 [Note : P(A/E1)=2/6 because if you get 1 after first roll, you have to get either 5 or 6 after second roll in order to make total sum at least 6. Similarly P(A/E2)=3/6 because if you get 2 after first roll,you can get 4 or 5 or 6 after second roll]<br />
<br />
=1/6+1/6 * 9/6= 5/12Probabilityhttps://gateoverflow.in/1751/gate2012_33?show=164088#a164088Mon, 30 Oct 2017 05:43:06 +0000Answered: Probability
https://gateoverflow.in/162709/probability?show=162809#a162809
<p>Audit records per day = 10 million = 10 * 10<sup>6 </sup>= 10<sup>7</sup></p>
<p>Records reflected by attacks per day = 10 * 20 = 200</p>
<p>Probability of attacks (intrusions) per day = $\frac{200}{10^{7}}$ = 2 * 10<sup>-5</sup></p>
<p>Probability of no intrusions = 1 - 2 * 10<sup>-5</sup> = 0.99998</p>
<p>Using Bayes theorem,</p>
<p>Required probability = $\frac{0.6*2*10^{-5}}{0.6*2*10^{-5} + 0.99998*0.0005}$ = 0.0234</p>
<p>So, only <strong>2.34 %</strong> of total alarms correspond to real intrusions.</p>Probabilityhttps://gateoverflow.in/162709/probability?show=162809#a162809Thu, 26 Oct 2017 04:06:24 +0000Median
https://gateoverflow.in/162030/median
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=16108385487096326986"></p>
<p> </p>Probabilityhttps://gateoverflow.in/162030/medianSun, 22 Oct 2017 16:04:12 +0000Answered: GATE 2012
https://gateoverflow.in/161893/gate-2012?show=161913#a161913
<p>We can solve this problem using Binomial distribution. Assuming that chocolates are distinguishable. Each chocolate has 1/3 probability to be given to the first child in the row.
<br>
<br>
P=$\frac{1}{3}$ q=1 - $\frac{1}{3}$ = $\frac{2}{3}$
<br>
<br>
<sup>10</sup>C<sub>3</sub>*$(1/3)^3 (2/3)^7$ = 5x$\frac{2^{10}}{3^9}$</p>Probabilityhttps://gateoverflow.in/161893/gate-2012?show=161913#a161913Sun, 22 Oct 2017 05:50:38 +0000Answered: Gate_other_branch_question
https://gateoverflow.in/161854/gate_other_branch_question?show=161901#a161901
<p>Here we have to know about the following property of mean and standard deviation :</p>
<blockquote>
<p>a) If we add something constant to each element of the current sample of data , then mean will change by the amount we changed each element . But standard deviation remains the same.</p>
</blockquote>
<p><em><strong>Hence D) should be the correct option. </strong></em></p>Probabilityhttps://gateoverflow.in/161854/gate_other_branch_question?show=161901#a161901Sun, 22 Oct 2017 05:23:51 +0000Answered: Gate_branch_question
https://gateoverflow.in/161856/gate_branch_question?show=161891#a161891
<p>The Gap is greater than 8 seconds, it means every next vehicle will arrive at least after 8 seconds.
<br>
We have to find the probability that no vehicle arrives in the gap of 8 seconds.
<br>
<br>
Expected vehicles per hour = 900/hour
<br>
<br>
λ = expected vehicles per 8 seconds = (900*8)/(60*60) = 2
<br>
<br>
λ = 2
<br>
<br>
P(X=x) = $(e^{-λ} *λ^x)$/x!
<br>
<br>
P(X=0) = $e^{-2}*2^0$/0! = e^-2 = <strong>1/e^2</strong> will be the answer.</p>
<p> </p>Probabilityhttps://gateoverflow.in/161856/gate_branch_question?show=161891#a161891Sun, 22 Oct 2017 03:33:38 +0000Answered: K. Rosen: Probability
https://gateoverflow.in/161474/k-rosen-probability?show=161487#a161487
<p><strong>It is given "3 appears twice as often as each other number".</strong> We can say from this statement</p>
<p>{1},{2},{4},{5},{6} are equally likely events therefore P(1) = P(2) = P(4) =P(5) =P(6) = (say,x)</p>
<p>while probability of event {3} = 2x</p>
<p>Sum of all probabilities = 1 = 2x + 5x</p>
<p>x = 1/7 Therefore P(1) = P(2) = P(4) =P(5) =P(6) = 1/7 while P(3) = 2/7</p>
<p>Probability that an odd number appears = P(1) + P(3) + P(5) = 1/7 + 2/7 + 1/7 = 4/7</p>Probabilityhttps://gateoverflow.in/161474/k-rosen-probability?show=161487#a161487Fri, 20 Oct 2017 04:58:27 +0000Answered: probability
https://gateoverflow.in/161146/probability?show=161174#a161174
i) all the five cards are spades with replacement=$\frac{13^{5}}{52^{5}}$<br />
<br />
ii) only 3 cards are spades =$\frac{13^{3}\times 39^{2}}{52^{5}}$<br />
<br />
iii) none is a spade=$\frac{39^{5}}{52^{5}}$Probabilityhttps://gateoverflow.in/161146/probability?show=161174#a161174Wed, 18 Oct 2017 18:41:24 +0000Answered: TIFR2010-A-19, TIFR2014-A-6
https://gateoverflow.in/18499/tifr2010-a-19-tifr2014-a-6?show=160855#a160855
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=13975540950192547316"></p>
<p><strong>Case 1</strong>:India Win that means Karan Lies and Arjun tells truth</p>
<p>In this case :- Because these are independent events so , P(Win) = P(Karan(Lie)) * P(Arjun(Truth)) = 2/3 * 3/4 *(1/3) = 1/6</p>
<p>Here notice that (1/3) is multiplied because we have selected one of three cases </p>
<p><strong>Case 2:</strong>India Lose means Karan tells truth and Arjun lies </p>
<p>Hence Probability for lose = > P(Lose ) = 1/3 * 1/4 *(1/3) = 1/36 </p>
<p><strong>Case 3</strong>:India draws the match in this case both telling lie</p>
<p>So probability = P(Draw) = P(Karan (Lie)) *P(Arjun(Lie)) = 2/3 * 1/4 *(1/3) = 1/18 </p>
<p>So according to bayes theorem = </p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=13879399395934761014"></p>
<p>Hence option B is right one</p>Probabilityhttps://gateoverflow.in/18499/tifr2010-a-19-tifr2014-a-6?show=160855#a160855Tue, 17 Oct 2017 16:10:51 +0000Answered: Gate_other_branch_question
https://gateoverflow.in/160738/gate_other_branch_question?show=160835#a160835
Answer : C 1/2<br />
<br />
we already know one is a boy. for both to be boys, younger one should also be boy. probability for younger one to be boy is 1/2 and that is total probability.Probabilityhttps://gateoverflow.in/160738/gate_other_branch_question?show=160835#a160835Tue, 17 Oct 2017 14:12:24 +0000Answered: GATE2016-2-05
https://gateoverflow.in/39541/gate2016-2-05?show=160819#a160819
<p>According to Baye's theorem :-</p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=7083944974035519647"></p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=12110418167166202448"></p>
<p>P(E) = 1.1/2 = 0.55</p>
<p> </p>Probabilityhttps://gateoverflow.in/39541/gate2016-2-05?show=160819#a160819Tue, 17 Oct 2017 12:05:48 +0000Answered: TIFR 2012- Probability
https://gateoverflow.in/120512/tifr-2012-probability?show=160806#a160806
"Amar says India has won", Anthony would actually consider 4 cases: <br />
<br />
India actually won: (Prob = 0.5) <br />
1. Amar tells truth, Akbar tells truth (Prob = 9/16) <br />
2. Amar lies, Akbar lies (Prob = 1/16) <br />
<br />
India actually loses: (Prob = 0.5) <br />
1. Amar tells truth, Akbar lies (Prob = 3/16) <br />
2. Amar lies, Akbar speaks truth (Prob = 3/16) <br />
<br />
Bayes theorem: <br />
(9/16 + 1/16) / ((9/16 + 1/16) + (3/16 + 3/16)) = 10/16Probabilityhttps://gateoverflow.in/120512/tifr-2012-probability?show=160806#a160806Tue, 17 Oct 2017 11:35:36 +0000Answered: Gate-2006, CE
https://gateoverflow.in/106695/gate-2006-ce?show=159879#a159879
Q. - > can we do it by both hyper geometric as well as by binomial distribution?<br />
<br />
No.<br />
<br />
Many people have given correct answer for this question. Now just mentioning another approach to get desired answer --><br />
<br />
We have to select 5 items in which 4 are of one type(non-defective ) and 1 is of another type (defective).<br />
<br />
$\Rightarrow \frac{5!}{4!\,1!} $ * (probability of selecting these items) $= \frac{5!}{4!\,1!} * \left ( \frac{2}{25}*\frac{23}{24}*\frac{22}{23}*\frac{21}{22}*\frac{20}{21} \right ) = \frac{\binom{2}{1}*\binom{23}{4}}{\binom{25}{5}}$Probabilityhttps://gateoverflow.in/106695/gate-2006-ce?show=159879#a159879Sat, 14 Oct 2017 07:04:40 +0000Answered: GATE2003-60, ISRO2007-45
https://gateoverflow.in/948/gate2003-60-isro2007-45?show=159577#a159577
<p>$f_k(t)$ denote the probability density function of time taken to execute module (Here k = 1, 2). Means probability that module k will take 5 unit of time is given by $f_k(5)$.
<br>
<br>
Now let us assume total execution time is 4 unit then it's probability will be given by -->
<br>
<br>
$f_1(0)*f_2(4)$ + $f_1(1)*f_2(3)$ + $f_1(2)*f_2(2)$ + $f_1(3)*f_2(1)$ + $f_1(4)*f_2(0)$
<br>
<br>
(means for example if first module takes 3 unit then another module should take 1 unit of time(and both probabilities are independent because module execution is sequential) so it's probability is $f_1(3)*f_2(1)$ ).
<br>
<br>
If entire program execution time is t unit then we can write general formula for it's probability as
<br>
<br>
$f_{total}(t) = $$\int_{0}^{t}$ $ f_1(x)*f_2(t-x)$dx$ $
<br>
<br>
which is also telling probability density at point 't' or $f_{total}(t)$ is our desired probability density function for entire program execution.</p>
<p> </p>
<p><strong>Answer is (C) part.</strong></p>Probabilityhttps://gateoverflow.in/948/gate2003-60-isro2007-45?show=159577#a159577Fri, 13 Oct 2017 02:01:00 +0000Answered: Rossen: Permutation and Combination
https://gateoverflow.in/159455/rossen-permutation-and-combination?show=159491#a159491
<p>Let's do this question by different approach:</p>
<p>Total permutations = n!</p>
<p>Part A: <strong>n precedes 1 and n −1 precedes 2</strong></p>
<p>By symmetery <strong>Event 1:</strong> half of n! permutations have n preceding 1 and other half has 1 preceding n ,so probability = 1/2</p>
<p>Similarly <strong>Event 2:</strong> half of n! permutations have n-1 preceding 2 and other half has 2 preceding n-1 ,so probability = 1/2</p>
<p>Both events occur together so probabilty = 1/4 (as both events are independent of each other)</p>
<p>Part B: <strong>n precedes 1 and 2</strong></p>
<p>Again By symmetery in total of n! permutations each of n or 1 or 2 can precede the other two(if n precede 1 and 2 then we don't care about how 1 and 2 are placed. We only care n precede both 1 and 2) therefore n precede both 1 and 2 in exactly($\frac{n!}{3}$) permutations so probability = 1/3</p>
<p> </p>Probabilityhttps://gateoverflow.in/159455/rossen-permutation-and-combination?show=159491#a159491Thu, 12 Oct 2017 12:48:59 +0000Answered: GATE 2011 probability
https://gateoverflow.in/33714/gate-2011-probability?show=159478#a159478
Answer is {n(n-1)/2}/{n^2} = (n-1)/(2*n)<br />
<br />
Here n = 6 so answer is 5/12.Probabilityhttps://gateoverflow.in/33714/gate-2011-probability?show=159478#a159478Thu, 12 Oct 2017 12:20:33 +0000Answered: Probability
https://gateoverflow.in/155533/probability?show=159470#a159470
<p>There are two cases to consider. One where the bus arrives before the train and the other way around. Representing the arrival times of the bus and the train as x and y coordinate respectively. We have the following equations representing the situation. </p>
<p>$y < x + t$ (i.e. train should arrive within t minutes after the bus.)</p>
<p>$x < y + 10$ (i.e. bus should arrive within 10 minutes after the train.)</p>
<p>We thus obtain the following plot:<img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=6317288224521468017"></p>
<p>The area of the required region represents time duration during which the train and the bus can meet. The sample space is represented by 60x60 plot (60 mins from 9 to 10). And we know that there is a 0.5 probability of the conditions being met. Thus, by dividing the area of the required region by total area of the sample space we get 0.5. </p>
<p>Area of the two triangles = $\frac{1}{2} *[(60-t)^2 + 50*50]$.</p>
<p>The sum of areas of the two triangles would equate to half the total area ($\because probability = 0.5$).</p>
<p>Thus we have:
<br>
=> $\frac{1}{2} *[(60-t)^2 + 50*50] = 3600/2 \\
<br>
= (60-t)^2 = 1100\\
<br>
= 60 - t = 10*\sqrt{11}\\
<br>
=> t = 26.83 mins$</p>Probabilityhttps://gateoverflow.in/155533/probability?show=159470#a159470Thu, 12 Oct 2017 11:59:34 +0000Answered: GATE1989-3-vii
https://gateoverflow.in/87141/gate1989-3-vii?show=158126#a158126
<p>A and C i think</p>
<p><a rel="nofollow" href="http://ocw.nctu.edu.tw/upload/classbfs121001554684839.pdf">http://ocw.nctu.edu.tw/upload/classbfs121001554684839.pdf</a></p>Probabilityhttps://gateoverflow.in/87141/gate1989-3-vii?show=158126#a158126Sat, 07 Oct 2017 13:21:07 +0000Answered: PROBABILITY
https://gateoverflow.in/157774/probability?show=157817#a157817
<p>This problem is an example of<em><strong> hypergeometric distribution.</strong></em>.It is similar to binomial distribution but differs with respect to following things :</p>
<blockquote>
<p>a) The population being considered here is finite and the mode of event is "without replacement".</p>
<p>b) The probability changes for each individual event , which violates the assumption of Binomial distribution that probabilty should remain the same.</p>
</blockquote>
<p>So here in total we have 10 balls , which is a finite population and out of which we have 4 red and 6 black balls..Now according to the question , favorable event is the one in which we have one red balls and two black balls..</p>
<p>Total number of events = Number of ways in which we can pick 3 out of 10 balls = <sup>10</sup>C<sub>3</sub></p>
<p>Now number of favorable events = Number of ways 1 red ball can be picked out of 4 red balls * Number of ways 2 black balls can be picked from 6 balls = <sup>4</sup>C<sub>1</sub> * <sup>6</sup>C<sub>2</sub></p>
<p>Hence required probability = ( <sup>4</sup>C<sub>1</sub> * <sup>6</sup>C<sub>2 </sub>) / <sup>10</sup>C<sub>3</sub></p>
<p><sub> </sub> = ( 4 * 15 * 6 ) / ( 10 * 9 * 8 )</p>
<p> <em><strong> = 1 / 2 </strong></em></p>
<p><em><strong>Ref : </strong></em><a rel="nofollow" href="https://en.wikipedia.org/wiki/Hypergeometric_distribution">https://en.wikipedia.org/wiki/Hypergeometric_distribution</a></p>Probabilityhttps://gateoverflow.in/157774/probability?show=157817#a157817Fri, 06 Oct 2017 10:11:45 +0000Answered: BINOMIAL DISTRIBUTION
https://gateoverflow.in/157752/binomial-distribution?show=157754#a157754
<p>0.2592</p>
<p>Unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice means $\frac{2}{6} = \frac{1}{3}$ is probability of each color comes on top.</p>
<p>Now ask <span class="marker">Atleast two</span></p>
<p>For 2= $\binom{3}{2}(\frac{1}{3})^{2} \frac{2}{3}$ = $\frac{4}{9}$</p>
<p>For 3= $\binom{3}{3}(\frac{1}{3})^{3}$ = $\frac{1}{27}$</p>
<p>Total = $\frac{7}{27}= 0.252$</p>Probabilityhttps://gateoverflow.in/157752/binomial-distribution?show=157754#a157754Fri, 06 Oct 2017 05:52:31 +0000test series
https://gateoverflow.in/157656/test-series
A fidget spinner has 3 wings. Each wing is numbered 1, 2,3. After spinning it vertically, the wing which is relatively upper than remaining is chosen. If two fidget spinners are spinned, the sum ‘r’ of the numbers turned up is considered. <br />
Choose the correct statements<br />
<br />
a)Probability (r≥4) = probability(r≤4)<br />
<br />
b)Probability (r≥4) = probability(r≤3)<br />
<br />
c)The probability(r>5) is greater than probability (r=4)<br />
<br />
d)All are falseProbabilityhttps://gateoverflow.in/157656/test-seriesThu, 05 Oct 2017 18:54:43 +0000Answered: GATE2014-1-48
https://gateoverflow.in/1927/gate2014-1-48?show=157405#a157405
Using Generating Functions<br />
<br />
$$x1+x2+x3+x4=22\ where\ 1\leq x_{i}\leq 6$$<br />
<br />
Now we need to find Cofficent of $x^{22}$<br />
$$=[X^{22}]\left \{ (x^{1}+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})^{4} \right \} \\ =[X^{22}][x^{4}]\left \{ (1+x^{1}+x^{2}+x^{3}+x^{4}+x^{5})^{4} \right \} \\ =[X^{18}]\left \{ (1+x^{1}+x^{2}+x^{3}+x^{4}+x^{5})^{4} \right \} \\$$<br />
$$=[X^{18}]\left \{ (1+x^{1}+x^{2}+x^{3}+x^{4}+x^{5})^{4} \right \} \\ =[X^{18}]\left ( \frac{1-x^{6}}{1-x} \right )^{4}\ using\ GP\ Formula \\ =[X^{18}]\left ( \frac{1}{1-x} \right )^{4} \left ( 1-x^{6} \right )^{4} \\ =[X^{18}]\underbrace{\left [ \sum_{r=0}^{\bowtie } \binom{4+r-1}{r}x^{r}\right ]} \underbrace{\left [ \sum_{r=0}^{4}\binom{4}{r} (-x^{6})^{r} \right ]} \\$$<br />
IN First and Second Part we can put the values as<br />
$$r=18 \ r=0 \\ r=12 \ r=1 \\ r=6 \ r=2 \\ r=0 \ r=3$$<br />
$$=\binom{21}{18}x^{18}-\binom{15}{12}\binom{4}{1}x^{18}+\binom{9}{6}\binom{4}{2}x^{18}-\binom{3}{0}\binom{4}{3}x^{18} \\ =(1330-1820+504-4) x^{18} \\ =10x^{18}$$<br />
So there are 10 ways to get sum as 22.<br />
<br />
I know this method seems tedious and long, but it is mechanical and universal if used carefully.Probabilityhttps://gateoverflow.in/1927/gate2014-1-48?show=157405#a157405Thu, 05 Oct 2017 05:46:45 +0000Answered: GATE2006-IT-22
https://gateoverflow.in/3561/gate2006-it-22?show=156837#a156837
<p>Actually, this is Geometric distribution.Here N is Geometric r.v which represents no of tosses required for first head.</p>
<p>The expected value of a Geom r.v N = <strong>1/P</strong></p>
<p>P(Head) = p</p>
<p>So,</p>
<p><strong>E(N) = 1/p</strong></p>
<h2 style="background: rgb(238, 238, 238); border: 1px solid rgb(204, 204, 204); padding: 5px 10px;"><strong>The correct answer is (A)1/p</strong></h2>Probabilityhttps://gateoverflow.in/3561/gate2006-it-22?show=156837#a156837Mon, 02 Oct 2017 21:57:13 +0000Answered: Sheldon Ross chapter 4 Random variables Qno-72
https://gateoverflow.in/156385/sheldon-ross-chapter-4-random-variables-qno-72?show=156403#a156403
<p>Let us do for one value of 'i' and same can be done for other values of 'i' as well..Let us take i = 6 here..</p>
<p>So it means that the 6th game must result to a win in order to win the series..Hence we need to ensure 3 wins in preceding 5 matches..Hence now this problem reduces to binomial distribution B(5 , 0.6) as we are considering the preceding 5 matches(trials) and probability of individual trial = 0.6..</p>
<p>Hence </p>
<p>P(series is won by strong team) = P(3 wins in 5 matches) * P(win in 6th match)</p>
<p> = B(5,0.6) * 0.6</p>
<p> = <sup>5</sup>C<sub>3</sub> (0.6)<sup>3</sup> (0.4)<sup>2</sup> * 0.6</p>
<p> = 0.20736</p>
<p> <em><strong> = 0.21 (correct to 2 decimal places)</strong></em></p>
<p>Similarly we can find for i = 4 , 5 and 7 as well.</p>
<p>P(win a 2 - out of 3 series) = <sup>3</sup>C<sub>2</sub> (0.6)<sup>2</sup> (0.4)</p>
<p> <em><strong> = 0.432</strong></em></p>Probabilityhttps://gateoverflow.in/156385/sheldon-ross-chapter-4-random-variables-qno-72?show=156403#a156403Sat, 30 Sep 2017 17:52:11 +0000Probability
https://gateoverflow.in/155820/probability
Imagine you are given a bag of n balls. You are told that at least 10% of the balls are blue, and no more than 90% of the balls are red. Asymptotically (for large n) how many balls do you have to draw from the bag to see a blue ball with probability at least 2/3? (You can assume that the balls are drawn with replacement.)Probabilityhttps://gateoverflow.in/155820/probabilityThu, 28 Sep 2017 10:29:26 +0000