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 $p(1+q)=$

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Math

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2^{34}

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\sqrt{2}

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Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 6 years, 2 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)= ${\cfrac { { 4 }^{ 4 } }{ { 1 }0^{ 4 } }}$

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

Tanishq Varshney - 6 years, 2 months ago

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

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

A Former Brilliant Member - 6 years, 2 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

@Deepanshu Gupta – Thanks for replying , so I have to practice REE Papers too ! Life just becomes tougher and tougher :(

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 - 6 years, 2 months ago

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

I have explained it in the 3 lines above it

@Rohit Shah – ok i understood. thanx $\ddot \smile$

Ans3: $\frac{2-\sqrt3}{2}$

Mahimn Bhatt - 6 years, 2 months ago

Thats correct

can u post ur solution brief

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

@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

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

Thank you everyone for helping

Q2 class notes in coaching i think 5/10 to power n - 4/10 to power n

Madhukar Thalore - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$