GATE Overflow for GATE CSE - Recent questions and answers in Probability
https://gateoverflow.in/qa/mathematics/probability
Powered by Question2AnswerAnswered: Probability Distribution
https://gateoverflow.in/389366/probability-distribution?show=389369#a389369
P(X>=2) given X<=4 and λ = 1 :<br />
<br />
Poisson Distribution falls under Discrete Random Variable so considering all the discrete points in the range {2,3,4}.<br />
<br />
= P(X=2) + P(X=3) + P(X=4)<br />
<br />
= (e^(-1) * 1^2)/2! + (e^(-1) * 1^3)/3! + (e^(-1) * 1^4)/4!<br />
<br />
= (1/e) * (½ + 1/6 + 1/24)<br />
<br />
= 17/24e<br />
<br />
= 0.26058Probabilityhttps://gateoverflow.in/389366/probability-distribution?show=389369#a389369Fri, 25 Nov 2022 13:29:35 +0000Answered: Conditional Probability
https://gateoverflow.in/389295/conditional-probability?show=389321#a389321
$P(M1)$: Probability of the man being an English = $\frac{40}{100}$<br />
<br />
$P(M2)$: Probability of the man being an American = $\frac{60}{100}$<br />
<br />
$P(Vowel | M1)$: Probability of the random letter chosen is a vowel given that the man is English = $\frac{3}{6}$<br />
<br />
$P(Vowel | M2)$: Probability of the random letter chosen is a vowel given that the man is American = $\frac{2}{5}$<br />
<br />
$P(Vowel)$: Probability of the random letter chosen is a vowel $=$ $P(M1).P(Vowel | M1) + P(M2).P(Vowel|M2)$ $= (40/100)(3/6) + (60/100)(2/5)$<br />
<br />
By Bayes Theorem,<br />
$P(M1 | Vowel)$ = $\frac{P(M1).P(Vowel | M1)}{P(Vowel)}$ = $\frac{(40/100)(3/6)}{(40/100)(3/6) + (60/100)(2/5)}$ = $\frac{5}{11}$Probabilityhttps://gateoverflow.in/389295/conditional-probability?show=389321#a389321Fri, 25 Nov 2022 04:23:05 +0000Answered: TIFR CS & SS 2018 SS part Question 12
https://gateoverflow.in/389114/tifr-cs-%26-ss-2018-ss-part-question-12?show=389127#a389127
Lets assume<br />
<br />
C1: coin having both side Tails <br />
<br />
C2: coin having both side Head<br />
<br />
C3: coin having one side Head and other side tail<br />
<br />
$P(C1) = 3/10$ <br />
<br />
$P(C2) = 4/10$ <br />
<br />
$P(C3) = 3/10$<br />
<br />
$P(Tails) = P(C1).P(Tails|C1) + P(C2).P(Tails|C2) + P(C3).P(Tails|C3)$<br />
<br />
$= (3/10)(2/2) + (4/10)(0/2) + (3/10)(1/2)$<br />
<br />
<br />
<br />
Other side can have a head if and only if the picked coin is C3<br />
<br />
we can also frame the question as find the probability that tails was from coin C3<br />
<br />
So required probability = $\frac{(3/10)(1/2)}{(3/10)(2/2) + (4/10)(0/2) + (3/10)(1/2)}$ = $\frac{1}{3}$Probabilityhttps://gateoverflow.in/389114/tifr-cs-%26-ss-2018-ss-part-question-12?show=389127#a389127Tue, 22 Nov 2022 13:12:17 +0000Answered: TIFR CSE 2022 | Part A | Question: 2
https://gateoverflow.in/381950/tifr-cse-2022-part-a-question-2?show=388941#a388941
$P\, (\text{any pair share same birthday})$ = $365*(\frac{1}{365^2})$<br />
<br />
$E[(a,b)]=\begin{cases} 1 & \text{ if a and b share same birthday}\\ 0& \text{ otherwise } \end{cases}$<br />
<br />
$\text{There are}\;\binom{n}{2}\; \text{pairs in total.}$<br />
<br />
$\text{Now, according to the question}\;\; \frac{n(n-1)}{2}*\frac{365}{365^2}\geq 1$<br />
<br />
$\text{So, minimum value of}\; n\; \text{would be 28. (because, }28*27 = 756>730\text{)}$<br />
<br />
$\text{So, the answer is option }\textbf{B}\;.$Probabilityhttps://gateoverflow.in/381950/tifr-cse-2022-part-a-question-2?show=388941#a388941Sat, 19 Nov 2022 13:42:48 +0000Answered: TIFR CSE 2022 | Part A | Question: 7
https://gateoverflow.in/381945/tifr-cse-2022-part-a-question-7?show=385955#a385955
Probability of choosing a number k uniformly at random from {1,…,N−1} = $\frac{1}{N-1}$<br />
<br />
Number of way to choose k beads from N white beads = $\binom{N}{k}$<br />
<br />
Number of way in which k black beads can occur sequentially in the circle = N <br />
<br />
Required Probability = $\frac{1}{N-1}*\frac{N}{\binom{N}{1}} + \frac{1}{N-1}*\frac{N}{\binom{N}{2}} + .... + \frac{1}{N-1}*\frac{N}{\binom{N}{N-1}}$<br />
<br />
= $\frac{N}{N-1}\sum_{k = 1}^{N - 1}\frac{1}{\binom{N}{k}}$<br />
<br />
Answer : CProbabilityhttps://gateoverflow.in/381945/tifr-cse-2022-part-a-question-7?show=385955#a385955Sat, 22 Oct 2022 04:32:40 +0000Answered: GATE CSE 2017 Set 2 | Question: 31
https://gateoverflow.in/118373/gate-cse-2017-set-2-question-31?show=384214#a384214
<p><strong>This is the toughest question of <u>2017 GATE </u>exam but you can easily understand the answer like this….…..</strong></p>
<hr>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=1904598397198745052"></p>Probabilityhttps://gateoverflow.in/118373/gate-cse-2017-set-2-question-31?show=384214#a384214Fri, 30 Sep 2022 19:45:47 +0000Answered: GATE CSE 2014 Set 1 | Question: 48
https://gateoverflow.in/1927/gate-cse-2014-set-1-question-48?show=383349#a383349
Let, x1, x2, x3, x4 are the results of the four dice respectively.<br />
<br />
Here, x1 + x2 + x3 + x4 = 22 → eqn 1<br />
<br />
4 <= x1, x2, x3, x4 <= 6 → $a$<br />
<br />
[ if anyone of the dice shows less than 4, the sum of the four can’t be 22, :) ]<br />
<br />
Let,<br />
<br />
x1 = 6 – y1<br />
<br />
x2 = 6 – y2<br />
<br />
x3 = 6 – y3<br />
<br />
x4 = 6 – y4<br />
<br />
Therefore,<br />
<br />
y1 = 6 – x1<br />
<br />
y2 = 6 – x2<br />
<br />
y3 = 6 – x3<br />
<br />
y4 = 6 – x4<br />
<br />
[ Here, form a, 0 <= y1, y2, y3, y4 <= 2 ]<br />
<br />
putting x1, x2, x3, x4 in eqn 1 ---<br />
<br />
6 – y1 + 6 – y1 + 6 – y1 + 6 – y1 = 22<br />
<br />
y1 + y2 + y3+ y4 = 2 → eqn 2<br />
<br />
No. of solutions of eqn 2 = (n + r – 1 C r – 1)<br />
<br />
= 2 + 4 – 1 C 4 – 1<br />
<br />
= 5C2<br />
<br />
= 10<br />
<br />
Therefore, The probability that the sum of the results being 22 = 10 / ( 6 ^ 4)<br />
<br />
= 10 / 1296<br />
<br />
So, X = 10.Probabilityhttps://gateoverflow.in/1927/gate-cse-2014-set-1-question-48?show=383349#a383349Wed, 21 Sep 2022 09:04:51 +0000Answered: GATE CSE 2014 Set 1 | Question: 2
https://gateoverflow.in/1717/gate-cse-2014-set-1-question-2?show=383120#a383120
<p><strong>Answer = $0.25$</strong></p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=6241077138839668010"></p>
<p>Let $X$ = Length of Shorter Stick</p>
<p>$\implies 0<X<\frac{1}{2}\implies X$ follows Uniform Distribution over $(0,\frac{1}{2})$</p>
<p>We know, In uniform distribution, the mean (first <a rel="nofollow" href="https://en.wikipedia.org/wiki/Moment_(mathematics)" title="Moment (mathematics)">moment</a>) of the distribution is:</p>
<p>${\displaystyle E(X)={\frac {1}{2}}(b+a).}$ Link : <a rel="nofollow" href="https://en.wikipedia.org/wiki/Continuous_uniform_distribution#:~:text=The%20mean%20(first%20moment)%20of%20the%20distribution%20is%3A">Continuous uniform distribution - Wikipedia</a></p>
<p><strong>Hence,</strong> the expected length of the shorter stick is = $Mean$ = $E[X] = \frac{b+a}{2} = \frac{0+\frac{1}{2}}{2} = \frac{1}{4} = 0.25$</p>Probabilityhttps://gateoverflow.in/1717/gate-cse-2014-set-1-question-2?show=383120#a383120Sun, 18 Sep 2022 09:13:02 +0000Answered: GATE Overflow | Mock GATE | Test 1 | Question: 53
https://gateoverflow.in/285426/gate-overflow-mock-gate-test-1-question-53?show=382856#a382856
X – 5.5 / 4.2 = 8.5 – 6.6 / 2.3 = 8.9 <br />
<br />
Which is equivlaent to 9Probabilityhttps://gateoverflow.in/285426/gate-overflow-mock-gate-test-1-question-53?show=382856#a382856Tue, 13 Sep 2022 12:36:53 +0000Answered: A First Course In Probability, 9th Edition, Indian Version, Sheldon Ross, Chapter 3, Problems, 33.
https://gateoverflow.in/382652/course-probability-edition-version-sheldon-chapter-problems?show=382664#a382664
<p style="text-align:center"><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=16773380693695095881"></p>
<p>In above figure,</p>
<ul>
<li>$R$ means “It rains” and $\overline R$ means “It doesn’t rain.</li>
<li>$P$ means “It is predicted that today will rain” and $\overline P$ means “It is not predicted that today will rain”.</li>
<li>$C$ means “Travels by car” and $\overline C$ means “Doesn’t travel by car”.</li>
</ul>
<p>$\therefore P(\overline C | R) = \frac{P(\overline C \cap R)}{P(R)} = \frac{\frac{1}{2} * \frac{4}{5} * 0 + \frac{1}{2} * \frac{1}{5} * 0.5}{\frac{1}{2}} = 0.1$</p>Probabilityhttps://gateoverflow.in/382652/course-probability-edition-version-sheldon-chapter-problems?show=382664#a382664Sat, 10 Sep 2022 11:48:22 +0000Answered: TIFR CSE 2022 | Part A | Question: 15
https://gateoverflow.in/381937/tifr-cse-2022-part-a-question-15?show=382532#a382532
<p>(1) $\pi$ is the initial distribution of particle being at location $i$.</p>
<p>$\implies \pi(1) + \pi(2) + \pi(3) + … + \pi(n) = 1$</p>
<p>(2) (First Step) When particle is in location $i$, it can move to any position in {$1,2,…,i$} with uniform probability ie probability of particle going to location $k$ is $\frac{1}{i}$ where $k \le i$.</p>
<p>(3) (Second Step) When particle is in location $k$ after first step, it can move to any position in {$k,k+1,…,n$} with uniform probability ie probability of particle going to location $j$ is $\frac{1}{n-k+1}$ where $j \ge k$.</p>
<p>As given, after completing both steps, the final distribution of particle is uniform in {$1,2,…,n$} with probability that particle is in location $j$ is $\pi’(j) = \frac{1}{n}$.</p>
<p>$\therefore \pi’(1) = \frac{1}{n}$</p>
<p>Now, the probability in terms of $\pi(i)$ of particle being at location $1$ is given by – </p>
<p>$\pi’(1) = \pi(1) * \frac{1}{1} * \frac{1}{n} + \pi(2) * \frac{1}{2} * \frac{1}{n} + \pi(3) * \frac{1}{3} * \frac{1}{n} + … + \pi(n) * \frac{1}{n} * \frac{1}{n}$</p>
<p>$\implies \frac{1}{n} = \pi(1) * \frac{1}{1} * \frac{1}{n} + \pi(2) * \frac{1}{2} * \frac{1}{n} + \pi(3) * \frac{1}{3} * \frac{1}{n} + … + \pi(n) * \frac{1}{n} * \frac{1}{n}$</p>
<p>$\implies 1 = \pi(1) * \frac{1}{1} + \pi(2) * \frac{1}{2} + \pi(3) * \frac{1}{3} + … + \pi(n) * \frac{1}{n}$</p>
<p>$\therefore \pi(1) + \pi(2) + \pi(3) + … + \pi(n) = \pi(1) * \frac{1}{1} + \pi(2) * \frac{1}{2} + \pi(3) * \frac{1}{3} + … + \pi(n) * \frac{1}{n}$</p>
<p>$\implies \pi(2) + \pi(3) + … + \pi(n) = \frac{\pi(2) }{2} + \frac{\pi(3)}{3} + … + \frac{\pi(n)}{n}$</p>
<p>This is only possible when $\pi(2) = \pi(3) = … = \pi(n) = 0$</p>
<p>$\implies \pi(1) = 1 \text{ and } \pi(i) = 0, \forall i \neq 1$</p>
<p><strong>Answer :- D.</strong></p>Probabilityhttps://gateoverflow.in/381937/tifr-cse-2022-part-a-question-15?show=382532#a382532Fri, 09 Sep 2022 06:05:20 +0000Answered: TIFR CSE 2022 | Part A | Question: 10
https://gateoverflow.in/381942/tifr-cse-2022-part-a-question-10?show=382418#a382418
<p>The first marble can be picked out in $n$ ways and the second one in $n-1$ ways, the total number of ways of picking $2$ marbles will be $n(n-1)$ ways.<br>
<br>
For getting 2 marble of the same colour the first one can be selected in $\binom{2m}{1}$ way and the second marble can be selected in only $1$ way(the same colour as the first draw), the total number of ways of picking 2 marble of the same colour will be $2m×1 = 2m$ ways.<br>
<br>
<strong>Probability that both marbles are of same colour</strong> = $2m/n(n-1)$</p>
<p>Hence, option <strong><em>C </em></strong>is the correct answer.</p>Probabilityhttps://gateoverflow.in/381942/tifr-cse-2022-part-a-question-10?show=382418#a382418Wed, 07 Sep 2022 09:17:38 +0000Answered: TIFR CSE 2022 | Part A | Question: 12
https://gateoverflow.in/381940/tifr-cse-2022-part-a-question-12?show=382420#a382420
<p>Either the box picked by Alice at the first attempt may have had the prize or may not have had the prize, there are two event here.<br>
<br>
Event $E1$: Alice picked the box with the prize on the first attempt which has a probability of $3/7$.</p>
<p>Now on the second attempt, she has a $2/5$ probability of winning the prize.<br>
<br>
Event $E2$: Alice didn't pick the box with the prize on the first attempt which has a probability of $4/7$.</p>
<p>Now on the second attempt, she has a $3/5$ probability of winning the prize.<br>
<br>
The probability of winning prize $P(W)$ is given as</p>
<p>$P(W) = P(E1)P(W/E1) + P(E2)P(W/E2)$</p>
<p> $=\left ( 3/7\right )*\left(2/5\right) + \left ( 4/7 \right )*\left ( 3/5 \right )$ = $18/35$<br>
<br>
Hence, option <em><strong>D</strong></em> is the correct answer.</p>Probabilityhttps://gateoverflow.in/381940/tifr-cse-2022-part-a-question-12?show=382420#a382420Wed, 07 Sep 2022 09:17:37 +0000Answered: GATE CSE 2021 Set 2 | Question: 29
https://gateoverflow.in/357511/gate-cse-2021-set-2-question-29?show=382144#a382144
<p>We can see here 2 cases, </p>
<ol style="list-style-type:decimal" type="1">
<li><strong>If she attempt 1st Qa , then Qb </strong></li>
</ol>
<ol start="2" style="list-style-type:decimal" type="1">
<li><strong>If she attempt 1st Qb and then Qa</strong></li>
</ol>
<p><strong>Qa = Attempt right , Qa’ = incorrect attempt ;</strong></p>
<p><strong>Qb = Attempt right, Qb’ = incorrect attempt</strong></p>
<p>Now we see this two cases by tree method ----</p>
<p><u><strong>Case 1:</strong></u></p>
<p><u><strong><img alt="" height="134" src="https://gateoverflow.in/?qa=blob&qa_blobid=17231915432405249631" width="335"></strong></u></p>
<p>Marks for both correct = (10 + 20) = 30</p>
<p>Only for Qa correct = 10 marks</p>
<p>So, Expected marks = (0.8 x 0.5 x 30) + (0.8 x 0.5 x 10) = <strong>16</strong></p>
<p><u><strong>Case 2:</strong></u></p>
<p><u><strong><img alt="" height="143" src="https://gateoverflow.in/?qa=blob&qa_blobid=6419892353303609617" width="358"></strong></u></p>
<p>Marks for both correct = (10 + 20) = 30</p>
<p>Only for Qb correct = 20 marks</p>
<p>So, Expected marks = (0.8 x 0.5 x 30) + (0.5 x 0.2 x 20) = <strong>14</strong></p>
<p><strong>So, we can make this conclusion that, <em>Case 1 expected marks > Case 2 expected marks</em></strong></p>
<p><strong><em>So, She should do first Qa and then Qb Expected marks 16.</em></strong></p>
<p><strong><em>Hence, option D is correct </em></strong></p>Probabilityhttps://gateoverflow.in/357511/gate-cse-2021-set-2-question-29?show=382144#a382144Sat, 03 Sep 2022 17:27:21 +0000Answered: TIFR CSE 2022 | Part A | Question: 5
https://gateoverflow.in/381947/tifr-cse-2022-part-a-question-5?show=382112#a382112
<p>Sample space will be Total no of functions = $m^{n}$</p>
<p>Since we need only the functions where f(x) = f(y) </p>
<p>so f(x) can map to any of the m elements which is same as selecting one element from m elements </p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=14261093207679200990"></p>Probabilityhttps://gateoverflow.in/381947/tifr-cse-2022-part-a-question-5?show=382112#a382112Sat, 03 Sep 2022 07:25:27 +0000maths probability
https://gateoverflow.in/380584/maths-probability
A dice is tossed seven times. What is the probability that all six faces appear at least once?Probabilityhttps://gateoverflow.in/380584/maths-probabilityMon, 15 Aug 2022 16:01:53 +0000A First Course In Probability, 9th Edition, Indian Version, Sheldon Ross, Chapter 2, Problems, 18.
https://gateoverflow.in/379369/course-probability-edition-version-sheldon-chapter-problems
A deck consists of 52 playing cards which is well shuffled. Draw 6 cards. Find the probability that among the cards there will be a representative of all suits?<br />
<br />
can someone get to this answer –-→ 6283420/20358520Probabilityhttps://gateoverflow.in/379369/course-probability-edition-version-sheldon-chapter-problemsThu, 28 Jul 2022 17:29:27 +0000maths
https://gateoverflow.in/378461/maths
A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card ?Probabilityhttps://gateoverflow.in/378461/mathsTue, 19 Jul 2022 14:10:26 +0000ISI Kolkata | MMA | 2022 --- A lie detector determines correctly ....
https://gateoverflow.in/376425/isi-kolkata-mma-2022-a-lie-detector-determines-correctly
<p><img alt="question" src="https://gateoverflow.in/?qa=blob&qa_blobid=8439975133878002557"></p>
<p>Picture of question</p>Probabilityhttps://gateoverflow.in/376425/isi-kolkata-mma-2022-a-lie-detector-determines-correctlyMon, 06 Jun 2022 18:23:48 +0000ISI2020-PCB-10
https://gateoverflow.in/375045/isi2020-pcb-10
Let $X_i\sim (i.i.d) Bernoulli(\frac{\lambda}{n}),n\geq\lambda\geq 0$. $Y_i\sim (i.i.d) Poisson(\frac{\lambda}{n})$, $\{X_i\}$ and $\{Y_i\}$ are independent. Let $\sum_{i=1}^{n^2} X_i=T_n$ and $\sum_{i=1}^{n^2} Y_i=S_n$ (say). Find the limiting distribution of $\frac{T_n}{S_n}$ as $n\to \infty$.Probabilityhttps://gateoverflow.in/375045/isi2020-pcb-10Fri, 06 May 2022 05:13:26 +0000Gate 2002 Question with a change.
https://gateoverflow.in/372648/gate-2002-question-with-a-change
<p>A box contains 10 screws, 3 of which are defective. Two screws are drawn at random <strong>with out replacement</strong>. The probability that none of the two screws is defective will be ..</p>
<p>This is a same question from Gate 2002, but it is having a small change, can anyone help me to find method to find it's answer.</p>
<p> </p>Probabilityhttps://gateoverflow.in/372648/gate-2002-question-with-a-changeTue, 08 Mar 2022 04:17:20 +0000TIFR 2011
https://gateoverflow.in/372372/tifr-2011
There are N boxes, each containing at most r balls. If the number of boxes containing at least i balls is $N_{i}$ for i = 1,2,...r, then the total number of balls contained in these N boxes<br />
<br />
(A) is exactly equal to $N_{1}$ +$N_{2}$ +...+$N_{r}$.<br />
<br />
(B) is strictly larger than $N_{1}$ +$N_{2}$ +...+$N_{r}$.<br />
<br />
(C) is strictly smaller than $N_{1}$ +$N_{2}$ +...+$N_{r}$.<br />
<br />
(D) cannot be determined from the given informationProbabilityhttps://gateoverflow.in/372372/tifr-2011Thu, 24 Feb 2022 09:13:35 +0000Probability
https://gateoverflow.in/371262/probability
<ol>
<li>A survey was conducted with recent MBAs of two different universities about their annual incomes. The following table displays data collected.</li>
</ol>
<table border="1">
<tbody>
<tr>
<td style="vertical-align:top; width:110.4pt">
<p>Annual Income</p>
</td>
<td style="vertical-align:top; width:111.3pt">
<p>University X</p>
</td>
<td style="vertical-align:top; width:111.35pt">
<p>University Y</p>
</td>
<td style="vertical-align:top; width:109.75pt">
<p>Total</p>
</td>
</tr>
<tr>
<td style="vertical-align:top; width:110.4pt">
<p>Under 8L</p>
</td>
<td style="vertical-align:top; width:111.3pt">
<p>36</p>
</td>
<td style="vertical-align:top; width:111.35pt">
<p>24</p>
</td>
<td style="vertical-align:top; width:109.75pt">
<p>60</p>
</td>
</tr>
<tr>
<td style="vertical-align:top; width:110.4pt">
<p>8L to 12L</p>
</td>
<td style="vertical-align:top; width:111.3pt">
<p>109</p>
</td>
<td style="vertical-align:top; width:111.35pt">
<p>56</p>
</td>
<td style="vertical-align:top; width:109.75pt">
<p>165</p>
</td>
</tr>
<tr>
<td style="vertical-align:top; width:110.4pt">
<p>12L and above</p>
</td>
<td style="vertical-align:top; width:111.3pt">
<p>35</p>
</td>
<td style="vertical-align:top; width:111.35pt">
<p>40</p>
</td>
<td style="vertical-align:top; width:109.75pt">
<p>75</p>
</td>
</tr>
<tr>
<td style="vertical-align:top; width:110.4pt">
<p>Total</p>
</td>
<td style="vertical-align:top; width:111.3pt">
<p>180</p>
</td>
<td style="vertical-align:top; width:111.35pt">
<p>120</p>
</td>
<td style="vertical-align:top; width:109.75pt">
<p>300</p>
</td>
</tr>
</tbody>
</table>
<p>Answer following questions using the table above</p>
<ol style="list-style-type:lower-alpha">
<li>What is the probability that a student is from University X if their salary is more than 12L?</li>
<li>What is the probability that an MBA gets salary less than 8L?</li>
<li>What is the probability that an MBA from university Y earns 10L annually?</li>
</ol>Probabilityhttps://gateoverflow.in/371262/probabilityTue, 08 Feb 2022 20:55:19 +0000Probability
https://gateoverflow.in/371261/probability
<ol>
<li>Use the following data.</li>
</ol>
<table border="1">
<tbody>
<tr>
<td style="vertical-align:top; width:108.75pt">
<p> </p>
</td>
<td style="vertical-align:top; width:105.75pt">
<p>Have pets</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>Do not have pets</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>Total</p>
</td>
</tr>
<tr>
<td style="vertical-align:top; width:108.75pt">
<p>Men</p>
</td>
<td style="vertical-align:top; width:105.75pt">
<p>0.41</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>0.08</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>0.49</p>
</td>
</tr>
<tr>
<td style="vertical-align:top; width:108.75pt">
<p>Women</p>
</td>
<td style="vertical-align:top; width:105.75pt">
<p>0.45</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>0.06</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>0.51</p>
</td>
</tr>
<tr>
<td style="vertical-align:top; width:108.75pt">
<p>Total</p>
</td>
<td style="vertical-align:top; width:105.75pt">
<p>0.86</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>0.14</p>
</td>
<td style="vertical-align:top; width:105.8pt">
<p>1</p>
</td>
</tr>
</tbody>
</table>
<ol style="list-style-type:lower-alpha">
<li>What is the probability of a randomly selected person being a man, given that they do not own a pet?</li>
<li>What is the probability for a woman to be a pet owner?</li>
<li>How likely is that a randomly selected person is a pet owner man?</li>
</ol>Probabilityhttps://gateoverflow.in/371261/probabilityTue, 08 Feb 2022 20:54:44 +0000Applied Grand Test 10
https://gateoverflow.in/370712/applied-grand-test-10
<div class="col-12 col-md-12 mb-2 p-0" style="width:1110px; padding:0px !important; max-width:100%; text-align:left; text-indent:0px; -webkit-text-stroke-width:0px">The Netherlands is one of the world leaders in the production and sale of tulips. Suppose the heights of the tulips in the green house of rotterdams fantastic flora follow a continuous uniform distribution with lower bound of 7 inches and upper bound of 16 inches. You have come to the greenhouse to select a bouquet of tulips, but only tulips with a height greater than 10 inches may be selected. What is the probability that a randomly selected tulip is tall enough to pick ____</div>Probabilityhttps://gateoverflow.in/370712/applied-grand-test-10Thu, 27 Jan 2022 09:38:57 +0000GATE Applied Mock Test
https://gateoverflow.in/370522/gate-applied-mock-test
<p><img alt="Question" src="https://gateoverflow.in/?qa=blob&qa_blobid=7230112558121014457"></p>
<ol style="list-style-type:upper-alpha" type="A">
<li>3</li>
</ol>
<ol start="2" style="list-style-type:upper-alpha" type="A">
<li>-1</li>
</ol>
<ol start="100" style="list-style-type:upper-roman" type="I">
<li>1</li>
</ol>
<ol start="500" style="list-style-type:upper-roman" type="I">
<li>2</li>
</ol>Probabilityhttps://gateoverflow.in/370522/gate-applied-mock-testMon, 24 Jan 2022 06:26:21 +0000Applied Mock Test
https://gateoverflow.in/368545/applied-mock-test
Among individuals who have sustained head injuries, x-rays reveal that only about 6% have skull fractures. Nausea is a standard symptom of a skull fracture and occurs in 98% of all skull fracture cases with other types of head injuries, nausea is present in about 70% of all cases. Suppose that an individual who has just suffered a head injury is not nauseous. Find the probability that he has a skull fracture _____Probabilityhttps://gateoverflow.in/368545/applied-mock-testWed, 29 Dec 2021 08:33:12 +0000if a 3 digit number
https://gateoverflow.in/368418/if-a-3-digit-number
if a 3 digit number is randomly chosen what is the probability that either the number or some permutation of the number (which is a 3 digit number ) is divisible by 4 and 5Probabilityhttps://gateoverflow.in/368418/if-a-3-digit-numberMon, 27 Dec 2021 15:52:06 +0000Conditional Probability
https://gateoverflow.in/368275/conditional-probability
<p style="text-align:center"><img alt="Conditional Probability" height="260" src="https://gateoverflow.in/?qa=blob&qa_blobid=15763123621577054698" width="450"></p>
<p>The above-given problem is from MIT-OCW probability – <a href="https://tinyurl.com/25wamnxz" rel="nofollow">https://tinyurl.com/25wamnxz</a></p>
<p> </p>
<p><strong>There are two biased coins – A and B. The probability of choosing either coin is 0.5. Once the coins are chosen, we perform the experiment of tossing the coins as shown in the figure above. The probabilities for heads and tails for respective coins are given below:</strong></p>
<p><strong>Coin A –</strong></p>
<p><strong> Probability of Head: 0.9</strong></p>
<p><strong> Probability of Tail: 0.1</strong></p>
<p><strong>Coin B –</strong></p>
<p><strong> Probability of Head: 0.1</strong></p>
<p><strong> Probability of Tail: 0.9</strong></p>
<p> </p>
<p>************************************************************************************************</p>
<p> </p>
<p><em><strong>Event A: Coin A is chosen</strong></em></p>
<p><em><strong>Event B: first 10 tosses are heads</strong></em></p>
<h3>What is the probability of <em><strong>(A | B)</strong></em> i.e. <strong>if we get 10 successive heads, then what is the probability that coin A was chosen?</strong></h3>Probabilityhttps://gateoverflow.in/368275/conditional-probabilitySun, 26 Dec 2021 06:49:34 +0000Testbook Test Series
https://gateoverflow.in/367754/testbook-test-series
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=17671914141581969313"></p>Probabilityhttps://gateoverflow.in/367754/testbook-test-seriesSat, 18 Dec 2021 09:55:41 +0000Testbook Quiz
https://gateoverflow.in/367490/testbook-quiz
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=16368737457206027201"></p>Probabilityhttps://gateoverflow.in/367490/testbook-quizTue, 14 Dec 2021 10:46:07 +0000Testbook Quiz
https://gateoverflow.in/367489/testbook-quiz
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=449063453069983912"></p>Probabilityhttps://gateoverflow.in/367489/testbook-quizTue, 14 Dec 2021 10:44:37 +0000Testbook Quiz
https://gateoverflow.in/367488/testbook-quiz
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=9663444196096831193"></p>Probabilityhttps://gateoverflow.in/367488/testbook-quizTue, 14 Dec 2021 10:43:52 +0000Testbook Quiz
https://gateoverflow.in/367487/testbook-quiz
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=5999015435077446038"></p>Probabilityhttps://gateoverflow.in/367487/testbook-quizTue, 14 Dec 2021 10:43:03 +0000CMI-2021-Data Science
https://gateoverflow.in/367189/cmi-2021-data-science
<p>In the finale of the Indian Idol programme, Ladies Special, there are four women contestants — Arunima, Priyamani, Razi and Shriya. The organisers send all participants a list of 7 songs, asking each one to pick one song for the finale. All the four girls pick the same song, <em>Pyar Hua Chupkese</em> from the film 1942 A Love Story. To resolve the tie, two days before the finale, the organisers prepare 7 sheets of paper and fold them. Exactly one of the 7 songs from the list is written on each folded sheet and the folded sheets are identical in all other aspects. First Arunima is asked to pick one sheet out of 7, then Priyamani is asked to pick one sheet out of the remaining 6, then Razi is asked to pick one out of the remaining 5 and finally Shriya is asked to pick one out of the remaining 4.</p>
<p> </p>
<p>(a) What is the probability that the sheet chosen by Razi has the song <em>Pyar Hua Chupkese</em>?</p>
<p>(b) When Shriya is asked to choose a sheet, she argues that the three sheets chosen by the others should be opened, and if <em>Pyar Hua Chupkese</em> is not chosen by anyone, she should be allowed to pick that song. This is agreed to. What is the probability that Shriya gets to sing the song <em>Pyar Hua Chupkese</em>?</p>Probabilityhttps://gateoverflow.in/367189/cmi-2021-data-scienceThu, 09 Dec 2021 05:50:02 +0000CMI-2021-Data Science
https://gateoverflow.in/367186/cmi-2021-data-science
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=10503097150664865367"></p>Probabilityhttps://gateoverflow.in/367186/cmi-2021-data-scienceThu, 09 Dec 2021 05:44:33 +0000CMI-2021-Data Science
https://gateoverflow.in/367183/cmi-2021-data-science
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=59553247374345472"></p>Probabilityhttps://gateoverflow.in/367183/cmi-2021-data-scienceThu, 09 Dec 2021 05:39:20 +0000NPTEL Assignment Question
https://gateoverflow.in/366977/nptel-assignment-question
<p>in a class we ask the students their birthdays; one-by-one. continue till a date repeats.</p>
<p>(Assume that there are 365 dates.)</p>
<p>Let x := #students asked.</p>
<ol style="list-style-type:upper-alpha" type="A">
<li>What is P (x = k) =?, for $k \epsilon n$</li>
</ol>
<p> </p>
<p> </p>Probabilityhttps://gateoverflow.in/366977/nptel-assignment-questionTue, 07 Dec 2021 11:34:22 +0000NPTEL Assignment Question
https://gateoverflow.in/366762/nptel-assignment-question
The length of a door handle (in cm), manufactured by a factory, is normally distributed with µ = 6.0 and σ = 0.2 . The place to fix on the door can allow error in length up to 0.3 cm. What percentage of handles manufactured in the factory will be defective? How much should be the value of σ , so that the number of defectives in reduced to only 5%?Probabilityhttps://gateoverflow.in/366762/nptel-assignment-questionMon, 06 Dec 2021 08:29:31 +0000NPTEL Assignment Question
https://gateoverflow.in/366761/nptel-assignment-question
A survey of people in given region showed that 25% drank regularly. The probability of death due to liver disease, given that a person drank regularly, was 6 times the probability of death due to liver disease, given that a person did not drink regularly. The probability of death due to liver disease in the region is 0.005. If a person dies due to liver disease what is the probability that he/she drank regularly?Probabilityhttps://gateoverflow.in/366761/nptel-assignment-questionMon, 06 Dec 2021 08:27:53 +0000#GATECSE
https://gateoverflow.in/366530/%23gatecse
<p>Most of the online video materials that I’ve come across for Engineering Mathematics are mainly focused on question-solving.</p>
<p>They don’t clear out the basic concepts, without which it’s quite difficult to gain proficiency in problem-solving.</p>
<p><strong>What is the best textbook to follow for engineering mathematics with respect to GATE CS/IT branch?</strong></p>
<p><strong>or any other source where the basic concepts will get cleared.</strong></p>Probabilityhttps://gateoverflow.in/366530/%23gatecseWed, 01 Dec 2021 13:23:31 +0000Applied Test Series
https://gateoverflow.in/365349/applied-test-series
<div class="col-12 col-md-12 mb-2 p-0" style="width:1110px; padding:0px !important; max-width:100%; text-align:left; text-indent:0px; -webkit-text-stroke-width:0px">A final exam is a discrete mathematics course consisting of 20 true-false questions, each worth 3 points and 10 multiple choice questions, each worth 4 points. Suppose that the probability that a student gives a correct answer to a true-false question is 0.9 and the probability that student gives a correct answer to a multiple choice question is 0.8. What is student expected score on the final exam___</div>Probabilityhttps://gateoverflow.in/365349/applied-test-seriesSun, 07 Nov 2021 09:17:12 +0000Applied Test Series
https://gateoverflow.in/365283/applied-test-series
<div class="col-12 col-md-12 mb-2 p-0" style="width:1110px; padding:0px !important; max-width:100%; text-align:left; text-indent:0px; -webkit-text-stroke-width:0px">A large bowl contains 15 balls that are identical except for their colors 6 are colored red, 5 are colored blue and 4 are colored green. What is the probability that two balls selected at random from the bowl have the same color_____</div>Probabilityhttps://gateoverflow.in/365283/applied-test-seriesFri, 05 Nov 2021 03:23:58 +0000Differentiate between Conditional probability and Joint Probability
https://gateoverflow.in/364297/differentiate-between-conditional-probability-probability
<p>Hey all,</p>
<p>I have a confusion regarding when to use P(A∩B) or P(A/B). Like in following question:<a name="27611" title="" id="27611"></a></p>
<div class="qa-q-view-main">
<div class="qa-q-view-content">
<div>
<p>A group consists of equal no of men and women .of this grp 20% of men and 50% of women are unemployed .If a person is selected at random from this group ,the probability of the selected person being employed is</p>
<p> </p>
<p>here, how to understand that 50% of women unemployed is P(Unemployed/Women) and not P(Women ∩ Unemployed) ? Thanks</p>
</div>
</div>
</div>Probabilityhttps://gateoverflow.in/364297/differentiate-between-conditional-probability-probabilitySun, 03 Oct 2021 17:03:02 +0000TIFR CSE 2021 | Part A | Question: 4
https://gateoverflow.in/358964/tifr-cse-2021-part-a-question-4
<p>What is the probability that at least two out of four people have their birthdays in the same month, assuming their birthdays are uniformly distributed over the twelve months?</p>
<ol style="list-style-type:upper-alpha" type="A">
<li>$\frac{25}{48}$</li>
</ol>
<ol start="2" style="list-style-type:upper-alpha" type="A">
<li>$\frac{5}{8}$</li>
</ol>
<ol start="100" style="list-style-type:upper-roman" type="I">
<li>$\frac{5}{12}$</li>
</ol>
<ol start="500" style="list-style-type:upper-roman" type="I">
<li>$\frac{41}{96}$</li>
</ol>
<ol start="5" style="list-style-type:upper-alpha" type="A">
<li>$\frac{55}{96}$</li>
</ol>Probabilityhttps://gateoverflow.in/358964/tifr-cse-2021-part-a-question-4Thu, 25 Mar 2021 09:16:24 +0000TIFR CSE 2021 | Part A | Question: 10
https://gateoverflow.in/358958/tifr-cse-2021-part-a-question-10
<p>Lavanya and Ketak each flip a fair coin (i.e., both heads and tails have equal probability of appearing) $n$ times. What is the probability that Lavanya sees more heads than ketak?</p>
<p>In the following, the binomial coefficient $\binom{n}{k}$ counts the number of $k$-element subsets of an $n$-element set.</p>
<ol style="list-style-type:upper-alpha" type="A">
<li>$\frac{1}{2}$</li>
</ol>
<ol start="2" style="list-style-type:upper-alpha" type="A">
<li>$\frac{1}{2}\left ( 1-\sum_{i=0}^{n}\frac{\binom{n}{i}^{2}}{2^{2n}} \right )$</li>
</ol>
<ol start="100" style="list-style-type:upper-roman" type="I">
<li>$\frac{1}{2}\left ( 1-\sum_{i=0}^{n}\frac{\binom{n}{i}}{2^{2n}} \right )$</li>
</ol>
<ol start="500" style="list-style-type:upper-roman" type="I">
<li>$\frac{1}{2}\left ( 1-\frac{1}{2^{2n}} \right )$</li>
</ol>
<ol start="5" style="list-style-type:upper-alpha" type="A">
<li>$\sum_{i=0}^{n}\frac{\binom{n}{i}}{2^{n}}$</li>
</ol>Probabilityhttps://gateoverflow.in/358958/tifr-cse-2021-part-a-question-10Thu, 25 Mar 2021 09:16:20 +0000TIFR CSE 2021 | Part B | Question: 11
https://gateoverflow.in/358942/tifr-cse-2021-part-b-question-11
<p>Suppose we toss a fair coin (i.e., both beads and tails have equal probability of appearing) repeatedly until the first time by which at least $\textit{two}$ heads and at least $\textit{two}$ tails have appeared in the sequence of tosses made. What is the expected number of coin tosses that we would have to make?</p>
<ol style="list-style-type:upper-alpha" type="A">
<li>$8$</li>
<li>$4$</li>
<li>$5.5$</li>
<li>$7.5$</li>
<li>$4.5$</li>
</ol>Probabilityhttps://gateoverflow.in/358942/tifr-cse-2021-part-b-question-11Thu, 25 Mar 2021 09:14:56 +0000GATE CSE 2021 Set 2 | Question: 22
https://gateoverflow.in/357518/gate-cse-2021-set-2-question-22
For a given biased coin, the probability that the outcome of a toss is a head is $0.4$. This coin is tossed $1,000$ times. Let $X$ denote the random variable whose value is the number of times that head appeared in these $1,000$ tosses. The standard deviation of $X$ (rounded to $2$ decimal place) is _________Probabilityhttps://gateoverflow.in/357518/gate-cse-2021-set-2-question-22Thu, 18 Feb 2021 09:16:41 +0000GATE CSE 2021 Set 2 | Question: 33
https://gateoverflow.in/357507/gate-cse-2021-set-2-question-33
<p>A bag has $r$ red balls and $b$ black balls. All balls are identical except for their colours. In a trial, a ball is randomly drawn from the bag, its colour is noted and the ball is placed back into the bag along with another ball of the same colour. Note that the number of balls in the bag will increase by one, after the trial. A sequence of four such trials is conducted. Which one of the following choices gives the probability of drawing a red ball in the fourth trial?</p>
<ol style="list-style-type:upper-alpha" type="A">
<li>$\dfrac{r}{r+b} \\$</li>
<li>$\dfrac{r}{r+b+3}\\$</li>
<li>$\dfrac{r+3}{r+b+3} \\$</li>
<li>$\left( \dfrac{r}{r+b} \right) \left ( \dfrac{r+1}{r+b+1} \right) \left( \dfrac{r+2}{r+b+2} \right) \left( \dfrac{r+3}{r+b+3} \right)$</li>
</ol>Probabilityhttps://gateoverflow.in/357507/gate-cse-2021-set-2-question-33Thu, 18 Feb 2021 09:15:48 +0000GATE CSE 2021 Set 1 | Question: 18
https://gateoverflow.in/357433/gate-cse-2021-set-1-question-18
The lifetime of a component of a certain type is a random variable whose probability density function is exponentially distributed with parameter $2$. For a randomly picked component of this type, the probability that its lifetime exceeds the expected lifetime (rounded to $2$ decimal places) is ____________.Probabilityhttps://gateoverflow.in/357433/gate-cse-2021-set-1-question-18Thu, 18 Feb 2021 08:52:10 +0000