IIT Kanpur Test Sample Paper

Question

(Batman and Robin)  The city of Gotham keeps propping  up newer challenges for our dynamic  duo.  The criminals  in the city are so organized  that they  have come up with a schedule to commit  the crimes.

a.  The Mad Hatter commits  maddeningly despicable  crimes from 10AM - 11AM

b.  The Riddler  does nasty  things  (that are hard  to understand) as well from 11AM - 12PM

c. 12PM to 1PM is lunch hour - no crimes get committed then

d.  The Scarecrow scares the living lights out of citizens of Gotham from 1PM - 2PM

e. The Bane makes a grand  appearance from 2PM - 3PM

f. The Joker  arrives  fresh from his afternoon  nap and wreaks havoc from 3PM - 4PM

g. Just  as the evening is setting,  The Catwoman commits  thefts  from 4PM - 5PM

To keep up with this busy schedule, Batman and Robin (separately and) independently choose a random shift and  they  go and defeat  whichever  criminal  is operating  in that shift.  Note that each chooses just one shift.  If they  happen  to choose the  lunch hour then  they  don’t fight anyone  and  just  go home - its a holiday  for them!

1) What is the probability that at least one of Batman and Robin gets a holiday?                                      

2) Bane  is too powerful  an  adversary - to  defeat  him,  Batman and  Robin  need to  fight  together.  All other  criminals  can  be defeated  by  Batman or Robin  alone. What  is the  probability that Bane  gets defeated?        

3) Batman has  a crush  on Catwoman and  would  like to  (ahem  ahem)  fight  her  alone.   What  is the probability of that?  

4) In the above scenario, what is the probability that two criminals get defeated?  

Answer 1

Total number of shift = 7
And the probability of each shift being chosen = $\frac{1}{7}$

Now as Batman and Robin separately and independently choose shift so $P(B \cap R)$ = $P(B).P(R)$

1. Batman choose lunch hour + Robin choose lunch hour - Both choose lunch hour.
 $\frac{1}{7}$ + $\frac{1}{7}$ - $\frac{1}{7}.\frac{1}{7}$ = $\frac{13}{49}$

2.  Bane gets defeated when both choose 2PM - 3PM shift
$\frac{1}{7}.\frac{1}{7}$ = $\frac{1}{49}$

3.  Batman chooses 4PM - 5PM shift and Robin choose any other shift.
$\frac{1}{7}.\frac{6}{7}$ = $\frac{6}{49}$

4. Both choose different shifts and shifts of Bane appearance and lunch are not choose.
$\frac{5}{7}.\frac{4}{6}$ = $\frac{10}{21}$

Comments

nice explanation

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For the 4th question, why denominator has 7*6?
Sample space would be choosing all possible combinations which is 7*7 isn't it?