probability

Question

the letters of the word PROBABILITY  are arranged in all possible ways . the chance that B's and also two I's occur together is .?

Comments

these type of question usually gate asks!!!  :)

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And generally we do commit mistakes( just by not understanding the question properly or misreading ) and end up getting {0, -0.33, -0.66 } marks. : )

Need to be careful.

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only this thing make gate difficult . otherwise it is not :)

Answer 1

Total no. of words possible $=\frac{11!}{2!2!}$ as it is a case of permuation with repetition.

Now, favorable case is when 2 B's and 2 I's occur together. So, this can be counted by ignoring 1 B and 1 I and then the number of words possible $={9!}$

So, Required Probability $=\frac{9!.2!2!}{11!} = \frac{2}{55}$

Comments

@arjun sir,i did'nt get this part b--

 

"Now, favorable case is when 2 B's and 2 I's occur together. So, this can be counted by ignoring 1 B and 1 I and then the number of words possible =9!"

whats wrong with 8!/2! 2! ??

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Here it is possible BB at one place and II at another then also the condition of 2 consecutive Bs and 2 consecutive Is is still satisfied..

It is not necessary that BBII together as a unit..They may be separate..

If the question were :

2 consecutive Bs followed by 2 consecutive As

 then ur answer is fine..

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oh..i thought the question is 2B and 2 I should occur tegether.

thanks for clarifying.