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http://gateoverflow.in/qa/mathematics/probability
Powered by Question2AnswerAnswered: 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: GATE2002-2.16
http://gateoverflow.in/846/gate2002-2-16?show=126096#a126096
<p>Total outcomes - 2<sup>4 </sup>(Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)
<br>
Now out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.</p>
<p>So, probability, favourable events/total outcome</p>
<p>14/16 = 7/8</p>Probabilityhttp://gateoverflow.in/846/gate2002-2-16?show=126096#a126096Tue, 18 Apr 2017 15:38:27 +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 +0000Answered: probability
http://gateoverflow.in/121747/probability?show=121764#a121764
2 white and 1 black balls can be obtained in following scenarios:<br />
<br />
White from first, White from second and Black from third OR White from first, Black from second and white from third OR Black from first, White from second and White from third:<br />
<br />
$Prob(\text{2 Whites and 1 Black ball}) = \frac{3}{6}*\frac{2}{4}*\frac{3}{4} + \frac{3}{6}*\frac{2}{4}*\frac{1}{4} + \frac{3}{6}*\frac{2}{4}*\frac{1}{4}$<br />
<br />
$ = \frac{5}{16}$Probabilityhttp://gateoverflow.in/121747/probability?show=121764#a121764Thu, 16 Mar 2017 12:31:49 +0000Answered: probability
http://gateoverflow.in/121744/probability?show=121759#a121759
Total machines: 4<br />
<br />
Total faulty machines: 2<br />
<br />
Total Non faulty machines : 2<br />
<br />
Only two test are needed: Both faulty machines come up in the first two checks OR Both non faulty machines come up in the first two checks<br />
<br />
$Prob(\text{Only two checks are needed}) = \frac{2}{4}*\frac{1}{3}+\frac{2}{4}*\frac{1}{3}$<br />
<br />
$ = \frac{1}{6}+\frac{1}{6}$<br />
<br />
$ = \frac{1}{3}$Probabilityhttp://gateoverflow.in/121744/probability?show=121759#a121759Thu, 16 Mar 2017 12:12:29 +0000Answered: probability
http://gateoverflow.in/121746/probability?show=121757#a121757
<p>Now n-persons are sitting in a row so we can select two people who are sitting consecutively in n-1 ways. Total number of ways of selecting two people is <sup>n</sup>C<sub>2.</sub></p>
<p>So the required probability is</p>
<p>( <sup>n</sup>C<sub>2</sub> - (n-1) ) / <sup>n</sup>C<sub>2</sub></p>Probabilityhttp://gateoverflow.in/121746/probability?show=121757#a121757Thu, 16 Mar 2017 12:07:36 +0000Answered: MEBook
http://gateoverflow.in/121682/mebook?show=121706#a121706
1.when we get 6 on the first throw =1/6<br />
<br />
2.getting 1on the first throw,we need 5 or 6 on the next throw.so 1/6* 1/6 + 1/6*1/6<br />
<br />
3.getting 2 on first throw,we need 4 or 5 or 6 on next throw,so 1/6* 1/6 + 1/6* 1/6 + 1/6* 1/6<br />
<br />
4.getting 3 on first throw,we need 3 or 4 or 5 or 6 on next throw,so 1/6* 1/6 + 1/6* 1/6 + 1/6* 1/6 + 1/6* 1/6<br />
<br />
total probability = 1/6 + 2*1/6* 1/6 + 3*1/6* 1/6 + 4*1/6* 1/6Probabilityhttp://gateoverflow.in/121682/mebook?show=121706#a121706Wed, 15 Mar 2017 20:35:39 +0000Answered: probability
http://gateoverflow.in/121543/probability?show=121545#a121545
6 forward 5 backward or 6 backward 5 forward<br />
<br />
11C6 * (0.4)^6 * (0.5)^5 + 11C6 * (0.4)^5 * (0.6)^5<br />
<br />
11C5 * (0.4)^5 * (0.6)^5 * (0.4 + 0.6)<br />
<br />
= 0.368 is the answerProbabilityhttp://gateoverflow.in/121543/probability?show=121545#a121545Tue, 14 Mar 2017 11:47:59 +0000Answered: ISRO 2013- probability [EE]
http://gateoverflow.in/121198/isro-2013-probability-ee?show=121264#a121264
<p>We can apply here BINOMIAL DISTRIBUTION,</p>
<p>Prob [defective] = 0.1</p>
<p>Prob [non defective] = 0.9</p>
<p>n = 10</p>
<p>r = 2 {exactly 2}</p>
<p>Binomial Distribution :-</p>
<p>nCr * (p)^r * (q)^(n-r)</p>
<p>10C2 * 0.1^2 * 0.9^8</p>
<p>0.1937102445 </p>
<p>or</p>
<p>0.2 </p>
<p><strong>Hence a) 0.2 is correct answer.</strong></p>
<p> </p>
<p> </p>Probabilityhttp://gateoverflow.in/121198/isro-2013-probability-ee?show=121264#a121264Sat, 11 Mar 2017 01:40:50 +0000Answered: ISRO 2013 Probability [EE]
http://gateoverflow.in/121197/isro-2013-probability-ee?show=121234#a121234
5C2+5C2/10C2=4/9 So option bProbabilityhttp://gateoverflow.in/121197/isro-2013-probability-ee?show=121234#a121234Fri, 10 Mar 2017 18:29:13 +0000Answered: ISRO 2014- probability [EE]
http://gateoverflow.in/121207/isro-2014-probability-ee?show=121222#a121222
<p>There are two possibilities,</p>
<p>Case 1: 4 out of 5 starting questions are answered then 6 out of remaining 8 questions. Possible number of ways is </p>
<p><sup>5</sup>C<sub>4 </sub>× <sup>8</sup>C<sub>6 </sub>= 140</p>
<p>Case 2: 5 out of 5 starting are answered and then 5 out of remaining 8. Possible number of ways is</p>
<p><sup>5</sup>C<sub>5 </sub>× <sup>8</sup>C<sub>5 </sub>= 56</p>
<p>So total number of ways is 140+56 = 196(option B)</p>Probabilityhttp://gateoverflow.in/121207/isro-2014-probability-ee?show=121222#a121222Fri, 10 Mar 2017 16:04:07 +0000Answered: ISRO 2014- Probability [Mech]
http://gateoverflow.in/120812/isro-2014-probability-mech?show=120939#a120939
.5*4/7+.5*.6=41/70.. So option dProbabilityhttp://gateoverflow.in/120812/isro-2014-probability-mech?show=120939#a120939Wed, 08 Mar 2017 14:04:55 +0000Answered: ISRO 2016 Probability [Mech]
http://gateoverflow.in/120796/isro-2016-probability-mech?show=120910#a120910
Answer is 0.6<br />
<br />
As selection of A and B are independent events<br />
<br />
P(A)*P(B)=0.3<br />
<br />
0.5*P(B)=0.3<br />
<br />
So P(B) is 0.6Probabilityhttp://gateoverflow.in/120796/isro-2016-probability-mech?show=120910#a120910Wed, 08 Mar 2017 11:18:30 +0000Answered: ISRO 2013- Probability [Mech]
http://gateoverflow.in/120891/isro-2013-probability-mech?show=120908#a120908
P(A)=.2<br />
<br />
P(B)=.3<br />
<br />
$P(A\cap B)=P(A)*P(B)=.2*.3=.06$<br />
<br />
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$<br />
<br />
$=.2+.3-.06$<br />
<br />
$=.44$Probabilityhttp://gateoverflow.in/120891/isro-2013-probability-mech?show=120908#a120908Wed, 08 Mar 2017 11:05:33 +0000ISRO 2013- Probability/Leap year [Mech]
http://gateoverflow.in/120893/isro-2013-probability-leap-year-mech
Chance that a leap year selected at random will contain 53 Sundays is<br />
(a) 3/7<br />
(b) 7/2<br />
(c) 7/3<br />
(d) 2/7Probabilityhttp://gateoverflow.in/120893/isro-2013-probability-leap-year-mechWed, 08 Mar 2017 09:15:25 +0000ISRO 2015- probability [Mech]
http://gateoverflow.in/120801/isro-2015-probability-mech
The probability that a teacher will give an unannounced test during any class is 1/5. If a student is absent twice, then probability that misses at least one test is<br />
<br />
(a) 24/25<br />
<br />
(b) 16/25<br />
<br />
(c) 7/25<br />
<br />
(d) 9/25Probabilityhttp://gateoverflow.in/120801/isro-2015-probability-mechTue, 07 Mar 2017 15:24:57 +0000ISRO 2016 Number of trials [Mech]
http://gateoverflow.in/120795/isro-2016-number-of-trials-mech
There are 20 locks and 20 matching keys. Maximum number of trials required to match all the locks is<br />
<br />
(a) 190<br />
<br />
(b) 210<br />
<br />
(c) 400<br />
<br />
(d) 40Probabilityhttp://gateoverflow.in/120795/isro-2016-number-of-trials-mechTue, 07 Mar 2017 15:01:05 +0000ISRO 2012- Probability [Mech]
http://gateoverflow.in/120790/isro-2012-probability-mech
The probability that A happens is 1/3. the odds against happening A are:<br />
<br />
a) 2:1<br />
<br />
b) 3:2<br />
<br />
c) 1:2<br />
<br />
d) 2:3Probabilityhttp://gateoverflow.in/120790/isro-2012-probability-mechTue, 07 Mar 2017 14:37:04 +0000ISRO 2012: MEch Probability
http://gateoverflow.in/120771/isro-2012-mech-probability
Let E and F be any two events with P(E U F)= 0.8, P(E) = 0.4 and P (E/ F) = 0.3. Then P (F) is <br />
<br />
(a) 3/7<br />
<br />
(b) 4/7<br />
(c) 3/5<br />
<br />
(d) 2/5Probabilityhttp://gateoverflow.in/120771/isro-2012-mech-probabilityTue, 07 Mar 2017 12:22:15 +0000Probability
http://gateoverflow.in/120760/probability
There are two Biased dice of which first dice shows an even number twice as frequently as odd number second dice shows 5, thrice as frequently is any other number. If these dice are rolled together what is the probability of getting---<br />
<br />
1)Sum as 10<br />
<br />
2)Sum more than 10.Probabilityhttp://gateoverflow.in/120760/probabilityTue, 07 Mar 2017 11:47:41 +0000TIFR 2012- Probability
http://gateoverflow.in/120512/tifr-2012-probability
<p>Amar and Akbar both tell the truth with probability 3/4 and lie with probability 1/4. Amar watches a test match and talks to Akbar about the outcome. Akbar, in turn, tells Anthony, "<strong>Amar told me that India won</strong>".</p>
<p>What probability should Anthony assign to India's win?</p>
<p>(a) 9/16</p>
<p>(b) 6/16</p>
<p>(c) 7/16</p>
<p>(d) 10/16</p>
<p> </p>Probabilityhttp://gateoverflow.in/120512/tifr-2012-probabilitySat, 04 Mar 2017 13:30:10 +0000