probability ques

Question

 

 

Comments

getting 4/15

Answer 1

Always in probability we can solve question atleast in 2 ways.

Method 1 :-  What exactly question asked

P( T₁ ⋂ T₂  ⋂ T₃ ⋂ T₄ ) = $\frac{no.\, of\, favorable\, cases }{Total\, possible \, ways}$

total possible ways = 5! =120

we need to analyses favorable cases for  T₁ ⋂ T₂  ⋂ T₃ ⋂ T₄

I) starts with 1  __   __  __  __

        second position should be 3 or 4 or 5

           i) 1 3 __ __ __  third position can't be 4 or 2 ==> should be 5 ==>  1 3 5 __ __   ==> fourth position should be 2 and fifth position should be 4 ===> 1 3 5 2 4

           ii) 1 4 __ __ __  third position can't be 3 or 5 ==> should be 2 ==>  1 4 2 __ __   ==> fourth position should be 3 and fifth position should be 5 ===> 1 4 2 5 3

         iii) 1 5 __ __ __  remaining is 2,3,4 but you have 3 positions only therefore you can't arrange

   total possible ways when starts with 1 = 2

II) starts with 2  __   __  __  __

        second position should be 4 or 5

           i) 2 4  __ __ __  third position can't be 3 or 5 ==> should be 1 ==>  2 4 1  __ __   ==> fourth position may be one of 3 or 5 and fifth position should be another one of 3 or 5 ===> 2 4 1 3 5 or 2 4 1 5 3

           ii) 2 5  __ __ __  third position can't be  4 ==> should be 3 or 1 (but if you take 1 next 3 and 4 are adjacent) ==>  2 5 3  __ __   ==> fourth position should be 1 and fifth position should be 4 ===> 2 5 3 1 4 

   total possible ways when starts with 2 = 3

III) starts with 3  __   __  __  __

        second position should be 5 or 1

           i)  3 1  __ __ __  third position can't be 2 ==> it may be 4 or 5 ==>  3 1 5 2 4  or 3 1 4 2 5

           ii) 3 5 __ __ __  third position can't be 4 ==> should be 1or 2 ==>  3 5 1 4 2 or  3 5 2 4 1

   total possible ways when starts with 3 = 4

IV) starts with 4  __   __  __  __

        second position should be 1 or 2

           i) 4 1  __ __ __  third position can't be 2 ==> may be 3 or 5 (if you take 5 at third position then fourth and fifth positions should be filled with 3 and 2 therefore should take 3 )==>  4 1 3  5 2

           ii) 4  2 __ __ __  third position can't be 1 or 3 ==> should be 5 ==>  4 2 5  __ __   ==> fourth position should be 3 or 1  ===> 4 2 5 3 1 or 4 2 5 1 3

   total possible ways when starts with 1 = 3

V) starts with 5  __   __  __  __

        second position should be 1 or 2 or 3

           i) 5 1  __ __ __  remaining (2,3,4) with three positions... you can't arrange with out either 2,3 or 3,4 are adjacent

           ii) 5 2 __ __ __  third position can't be  3 or 1 ==> should be 4  ==> fourth position should be 1 and fifth position should be 3 ===> 5 2 4 1 3

         iii) 5 3 __ __ __  third position can't be  2 or 4 ==> should be 1 ==>  5 3 1  __ __   ==> fourth position should be 4 and fifth position should be 2 ===> 5 3 1 4 2 

   total possible ways when starts with 2 = 2

Total favorable cases = 2+3+4+3+2 = 14

required probability = $\frac{14}{120} = \frac{7}{60}$

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P( T₁^' ⋂ T₂^'  ⋂ T₃^' ⋂ T₄^' )

Method 2 :- complement of the question 

P( T₁ ⋂ T₂  ⋂ T₃ ⋂ T₄ )  = 1 - P(( T₁ ⋂ T₂  ⋂ T₃ ⋂ T₄ )') = 1 - P( T₁^' ∪ T₂^'  ∪ T₃^' ∪ T₄^' )    where T_i^' is S_i and S_i+1 should be adjacent...

P( T₁^' ∪ T₂^'  ∪ T₃^' ∪ T₄^' ) = ( P( T₁^') +  P( T₂^') +  P( T₃^') +  P( T₄^')  ) - ( P( T₁^'⋂ T₂^'  ) + P( T₁^'⋂ T₃^' ) + P( T₁^'⋂ T₄^' ) + P( T₂^'⋂ T₃^') + P( T₂^' ⋂ T₄^' ) + P( T₃^' ⋂ T₄^' ) ) + ( P( T₁^' ⋂ T₂^'  ⋂ T₃^') + P( T₁^' ⋂ T₂^' ⋂ T₄^' ) + P( T₁^' ⋂ T₃^' ⋂ T₄^' ) + P(T₂^' ⋂ T₃^' ⋂ T₄^' ) ) - ( P( T₁^' ⋂ T₂^'  ⋂ T₃^' ⋂ T₄^' ))

= ($\frac{48}{120}+\frac{48}{120}+\frac{48}{120}+\frac{48}{120}) - ( \frac{12}{120}+\frac{24}{120}+\frac{24}{120}+\frac{12}{120}+\frac{24}{120}+\frac{12}{120} ) + ( \frac{4}{120}+ \frac{8}{120}+\frac{8}{120}+\frac{4}{120} ) - (\frac{2}{120})$

= $\frac{192-108+24-2}{120} $ = $\frac{106}{120}$

Required Probability = 1 - $\frac{106}{120}$ = $\frac{14}{120}$ = $\frac{7}{60}$

Comments

thanks for your effort.let me know it if there is any other simpler approach than these.

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in Method 1, you should find no.of favorable cases... 

for calculating no.of favorable cases 

                           1) you have to follow my approach

                           2) Direct formula ( may exist )

By Direct formula you can get answer simply but you have to remember the formula.... i strongly doesn't recommended this method [ it's my opinion ]

in Method 2, you can't simplify further... 

Answer 2

We can select and arrange 5 student in these ways $T_{1}={\color{Blue} 1}{\color{Pink} 4}{\color{Blue}2 }{\color{Pink} 5}{\color{Blue}3 }$=$\frac{3!2!}{5!}=\frac{12}{120}$

We can select and arrange 5 student in these ways$T_{2}={\color{Blue} 1}{\color{DarkOrange} 3}{\color{Orange}5 }{\color{Blue} 2}{\color{Pink} 4}$=$\frac{2!2!1!}{5!}=\frac{4}{120}$

We can select and arrange 5 student in these ways$T_{3}={\color{Blue} 2}{\color{DarkOrange} 4}{\color{Blue} 1}{\color{Pink} 3}{\color{Pink} 5}$=$\frac{2!2!1!}{5!}=\frac{4}{120}$

We can select and arrange 5 student in these ways$T_{4}={\color{Blue} 3}{\color{Pink} 1}{\color{Blue} 4}{\color{Pink} 2}{\color{Blue} 5}$=$\frac{3!2!}{5!}=\frac{12}{120}$

So, totally $\frac{32}{120}=\frac{4}{15}$ ways

Comments

ans is C

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@ srestha Mam, all those are intersection therefore you can't add them

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hmm

but there is no intersection,rt?

this link might be useful https://math.stackexchange.com/questions/96718/number-of-permutations-in-which-no-two-consecutive-numbers-are-adjacent

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All are dependent event ma'am.