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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, 6 months ago

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Solution-4) $$\displaystyle{P(A.B+A.C)=P(A.B)+P(A.C)-P((A.B).(A.C))\\ \\ P(A.B+A.C)=P(A.B)+P(A.C)-P(A.B.C)\\ P(A.B+A.C)=pq+\cfrac { p }{ 2 } -\cfrac { pq }{ 2 } =\cfrac { 1 }{ 2 } \\ \boxed { p(1+q)=1 } }$$ · 2 years, 6 months ago

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)=$${\cfrac { { 4 }^{ 4 } }{ { 1 }0^{ 4 } }}$$ · 2 years, 6 months ago

thanks, i didn't know that. why it shouldn't be $$\frac{4^4}{10^4}$$?? · 2 years, 6 months ago

Yes absolutely ! Sorry It was a typo ! Thanks I edited that · 2 years, 6 months ago

You actually remember the question from the 90s , dude your memory's also great !! · 2 years, 6 months ago

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 · 2 years, 6 months ago

Thanks for replying , so I have to practice REE Papers too ! Life just becomes tougher and tougher :( · 2 years, 6 months ago

For Q1) I guess answer is 3 $$P(Both\quad boys\quad and\quad girls\quad being\quad present)=1-P(only\quad boys\quad or\quad only\quad girls)\\ P(Only\quad boys)\quad =\quad \frac { 1 }{ { 2 }^{ n } } \\ Similarly,\\ P(Only\quad girls)\quad =\quad \frac { 1 }{ { 2 }^{ n } } \\ So\quad ,\\ P(A)\quad =\quad (1-\frac { 1 }{ 2^{ n-1 } } )\\ P(B)\quad =\quad P(Atmost\quad one\quad girl\quad being\quad present)\\ \quad \quad \quad \quad =\quad P(No\quad girl\quad is\quad present)\quad +\quad P(1\quad girl\quad is\quad present)\\ \quad \quad \quad \quad =\quad \frac { 1 }{ { 2 }^{ n } } +\frac { n }{ { 2 }^{ n } } \\ P(A\sqcap B)\quad =\quad P(1\quad girl\quad being\quad present)\\ \quad \quad \quad \quad \quad \quad \quad =\quad \frac { n }{ { 2 }^{ n } } \\ For\quad events\quad to\quad be\quad independent,\\ P(A).P(B)\quad =\quad P(A\sqcap B)\\ Solving\quad we\quad get\quad n\quad =\quad 3$$ · 2 years, 6 months ago

can u elaborate how u ended up with $$P(A)=1-\frac{1}{2^{n-1}}$$ · 2 years, 6 months ago

I have explained it in the 3 lines above it · 2 years, 6 months ago

ok i understood. thanx $$\ddot \smile$$ · 2 years, 6 months ago

Q2 class notes in coaching i think 5/10 to power n - 4/10 to power n · 2 years, 6 months ago

Thank you everyone for helping · 2 years, 6 months ago

$$3)\quad Answer\quad is\quad (1\quad -\quad \frac { \sqrt { 3 } }{ 2 } )\\ A\quad short\quad solution\quad :\quad Let\quad length\quad be\quad x\quad ,\quad breadth\quad will\quad be\quad (2l-x).\\ \quad Required\quad quadratic\quad ,\quad (x)(2l-x)<\frac { { l }^{ 2 } }{ 4 } \\ Solving\quad for\quad x\quad >\quad (\frac { 2+\sqrt { 3 } }{ 2 } )l\quad ,\quad x<(\frac { 2-\sqrt { 3 } }{ 2 } )l\\ Hence\quad the\quad required\quad probability$$ · 2 years, 6 months ago

Comment deleted Mar 23, 2015

i have the answer as $$p(1+q)=1$$ but i got my answer no where close to it. · 2 years, 6 months ago

What's the source of these questions ? · 2 years, 6 months ago

i have got practice papers and all these are a part of them · 2 years, 6 months ago

Ans3: $$\frac{2-\sqrt3}{2}$$ · 2 years, 6 months ago

can u post ur solution brief · 2 years, 6 months ago

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) · 2 years, 6 months ago

@Tanishq Varshney Kindly see the solution by rohit shah. I have done by the same method. · 2 years, 6 months ago

Thats correct · 2 years, 6 months ago