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Practice for you, a help for me

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 \(p(1+q)=\)

Note by Tanishq Varshney
2 years, 2 months ago

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Solution-4) \(\displaystyle{P(A.B+A.C)=P(A.B)+P(A.C)-P((A.B).(A.C))\\ \\ P(A.B+A.C)=P(A.B)+P(A.C)-P(A.B.C)\\ P(A.B+A.C)=pq+\cfrac { p }{ 2 } -\cfrac { pq }{ 2 } =\cfrac { 1 }{ 2 } \\ \boxed { p(1+q)=1 } }\) Deepanshu Gupta · 2 years, 2 months ago

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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)=\({\cfrac { { 4 }^{ 4 } }{ { 1 }0^{ 4 } }}\) Deepanshu Gupta · 2 years, 2 months ago

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@Deepanshu Gupta thanks, i didn't know that. why it shouldn't be \(\frac{4^4}{10^4}\)?? Tanishq Varshney · 2 years, 2 months ago

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@Tanishq Varshney Yes absolutely ! Sorry It was a typo ! Thanks I edited that Deepanshu Gupta · 2 years, 2 months ago

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@Deepanshu Gupta You actually remember the question from the 90s , dude your memory's also great !! Azhaghu Roopesh M · 2 years, 2 months ago

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@Azhaghu Roopesh M 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 Deepanshu Gupta · 2 years, 2 months ago

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@Deepanshu Gupta Thanks for replying , so I have to practice REE Papers too ! Life just becomes tougher and tougher :( Azhaghu Roopesh M · 2 years, 2 months ago

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

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@Rohit Shah can u elaborate how u ended up with \(P(A)=1-\frac{1}{2^{n-1}}\) Tanishq Varshney · 2 years, 2 months ago

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@Tanishq Varshney I have explained it in the 3 lines above it Rohit Shah · 2 years, 2 months ago

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@Rohit Shah ok i understood. thanx \(\ddot \smile\) Tanishq Varshney · 2 years, 2 months ago

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Q2 class notes in coaching i think 5/10 to power n - 4/10 to power n Madhukar Thalore · 2 years, 2 months ago

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Thank you everyone for helping Tanishq Varshney · 2 years, 2 months ago

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

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Comment deleted Mar 23, 2015

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@Rohit Shah i have the answer as \(p(1+q)=1\) but i got my answer no where close to it. Tanishq Varshney · 2 years, 2 months ago

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@Tanishq Varshney What's the source of these questions ? Rohit Shah · 2 years, 2 months ago

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@Rohit Shah i have got practice papers and all these are a part of them Tanishq Varshney · 2 years, 2 months ago

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Ans3: \(\frac{2-\sqrt3}{2}\) Mahimn Bhatt · 2 years, 2 months ago

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@Mahimn Bhatt can u post ur solution brief Tanishq Varshney · 2 years, 2 months ago

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@Tanishq Varshney 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) Rohit Shah · 2 years, 2 months ago

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@Rohit Shah @Tanishq Varshney Kindly see the solution by rohit shah. I have done by the same method. Mahimn Bhatt · 2 years, 2 months ago

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@Mahimn Bhatt Thats correct Rohit Shah · 2 years, 2 months ago

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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 Tanishq Varshney · 2 years, 2 months ago

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