The following are some probabality questions , do help by posting a solution..\(\ddot \smile\)

\(Q1)\) Consider families with \(n\) children and let \(A\) be the event that a family has children both **boys** and **girls** and \(B\) be the event that there is **at most one girl** in the family. Find the value of \(n\) for which the event \(A\) and \(B\) are **independent** ,assuming that each child has probability \(\frac{1}{2}\) of being a **boy**.

\(Q2)\) If \(four~whole~numbers\) taken at random are multiplied together, then the chance that the last digit of the product is \(1,3,7~or~9\) is.

\(Q3)\) A piece of wire of length \(4l\) is bent at random to form a rectangle, Find the **probability** that its area is **at most** \(\frac{l^2}{4}\)

finally,the last one

\(Q4)\) A person is assigned to \(3\) jobs \(A,B.~and~C\).The probability of his doing the jobs \(A,B~and~C\) are \(p,q~and~\frac{1}{2}\).He gets the full payment only if he either does job \(A\) and \(B\) or the job \(A\) and \(C\).If the probability of his getting the full payment is \(\frac{1}{2}\) then \(p(1+q)=\)

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## Comments

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TopNewestQ2 class notes in coaching i think 5/10 to power n - 4/10 to power n

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Thank you everyone for helping

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Solution-4) \(\displaystyle{P(A.B+A.C)=P(A.B)+P(A.C)-P((A.B).(A.C))\\ \\ P(A.B+A.C)=P(A.B)+P(A.C)-P(A.B.C)\\ P(A.B+A.C)=pq+\cfrac { p }{ 2 } -\cfrac { pq }{ 2 } =\cfrac { 1 }{ 2 } \\ \boxed { p(1+q)=1 } }\)

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\(3)\quad Answer\quad is\quad (1\quad -\quad \frac { \sqrt { 3 } }{ 2 } )\\ A\quad short\quad solution\quad :\quad Let\quad length\quad be\quad x\quad ,\quad breadth\quad will\quad be\quad (2l-x).\\ \quad Required\quad quadratic\quad ,\quad (x)(2l-x)<\frac { { l }^{ 2 } }{ 4 } \\ Solving\quad for\quad x\quad >\quad (\frac { 2+\sqrt { 3 } }{ 2 } )l\quad ,\quad x<(\frac { 2-\sqrt { 3 } }{ 2 } )l\\ Hence\quad the\quad required\quad probability\)

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Comment deleted Mar 23, 2015

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i have the answer as \(p(1+q)=1\) but i got my answer no where close to it.

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What's the source of these questions ?

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Solution-2)- It's actually Logical based question , In old 90's IIT Roorkee (REE) generally ask questions on this concept !Concept:- If n whole numbers (here n=4) are multiplied then the units place in the product will be 1 or 3 or 7 or 9 , if and only if , the units place of each of whole numbers is randomly selected also has 1 or 3 or 7 or 9 .\(Note\): This concepts also hold for :(B):: 1 or 3 or 5 or 7 or 9 and (C):: 1 or 2 or 3 or 4 or 6 or 7 or 8 or 9 .

Hence , P(units place of products of 4 numbers ends in 1 or 3 or 7 or 9)=\({\cfrac { { 4 }^{ 4 } }{ { 1 }0^{ 4 } }}\)

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You actually remember the question from the 90s , dude your memory's also great !!

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haha , never ! my meomry is too weak . I know this beacuse I had solved previous year REE papers(Maths only) , Recently . Since every one says that REE -Maths was tough ! And I realise it too :P

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thanks, i didn't know that. why it shouldn't be \(\frac{4^4}{10^4}\)??

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Yes absolutely ! Sorry It was a typo ! Thanks I edited that

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Ans3: \(\frac{2-\sqrt3}{2}\)

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can u post ur solution brief

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Assume length of one side to be x so other side would be 2l-x. Make a quadratic to get a range of values of x. Divide by all possible x that is (0,2l)

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@Tanishq Varshney Kindly see the solution by rohit shah. I have done by the same method.

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Thats correct

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For Q1) I guess answer is 3 \(P(Both\quad boys\quad and\quad girls\quad being\quad present)=1-P(only\quad boys\quad or\quad only\quad girls)\\ P(Only\quad boys)\quad =\quad \frac { 1 }{ { 2 }^{ n } } \\ Similarly,\\ P(Only\quad girls)\quad =\quad \frac { 1 }{ { 2 }^{ n } } \\ So\quad ,\\ P(A)\quad =\quad (1-\frac { 1 }{ 2^{ n-1 } } )\\ P(B)\quad =\quad P(Atmost\quad one\quad girl\quad being\quad present)\\ \quad \quad \quad \quad =\quad P(No\quad girl\quad is\quad present)\quad +\quad P(1\quad girl\quad is\quad present)\\ \quad \quad \quad \quad =\quad \frac { 1 }{ { 2 }^{ n } } +\frac { n }{ { 2 }^{ n } } \\ P(A\sqcap B)\quad =\quad P(1\quad girl\quad being\quad present)\\ \quad \quad \quad \quad \quad \quad \quad =\quad \frac { n }{ { 2 }^{ n } } \\ For\quad events\quad to\quad be\quad independent,\\ P(A).P(B)\quad =\quad P(A\sqcap B)\\ Solving\quad we\quad get\quad n\quad =\quad 3\)

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can u elaborate how u ended up with \(P(A)=1-\frac{1}{2^{n-1}}\)

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I have explained it in the 3 lines above it

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@Brian Charlesworth sir ,@Rohit Shah @Azhaghu Roopesh M ,@Deepanshu Gupta ,@Raghav Vaidyanathan @Calvin Lin sir, and all other brilliant members , do post a hint or a solution for above problems

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