Latest CTS Aptitude Question SOLUTION: there are 5 dogs and three cats. what is the probability that there is one cat at both the ends?

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(#M40038683) CTS Numerical Ability question Keep an EYE

there are 5 dogs and three cats. what is the probability that there is one cat at both the ends?

Asked In CTS Abitha (10 years ago)

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vikash kumar  (10 years ago)

(3C2*2!*6!)/8!=3/28

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Pranoy Bej  (10 years ago)

Let Dog= D and Cat= C
then the arrangement will be like CD_D_D_D_DC if their is Cat at both ends. Remaining 1 Cat cat be fit any of this gap. so 4C1 = 4
now 5dogs and 3cats can be arranged 6C3= 20
the probability P(E)=4/20 = 1/5

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Jagadeesh Tangadgi  (10 years ago)

@vikash kumar explain in detail

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Shivam Kumar Singh  (6 years ago)

suppose we have 8 places to fill and at 1st position and last position we have select 2 cats out of 3..
now at first place we will be selecting 1 cat out of 3=3C1 ..so ultimately at last we have to choose 1 cat out of 2 cats=2C1...now we are left with 5 dogs and 1 cat..possible way of arranging all 6 =6!..between two cars at extremeties...
total arrangements=8!..
so answer is 3C1*6!*2C1/8!=3/28

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