Latest Elitmus Aptitude Question SOLUTION: A Box Contain 3 Red And 2 White balls and balls are taken out one by one Without replacement. So what is the Probability the White Balls Always come at Last?

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(#M40034211) ELITMUS Numerical Ability question Probability Keep an EYE

A Box Contain 3 Red And 2 White balls and balls are taken out one by one Without replacement. So what is the Probability the White Balls Always come at Last?

Asked In Elitmus (10 years ago)

Solved ARPIT SRIVASTAVA Read Solution (11)

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Read Solution (11) :

sandeep  (10 years ago)

ans-4/10 i.e 2/5
total possibilities=5!/(3!*2!)=10
possibility that white is last = 4!/(3!*1!)=4
so,probability=4/10=2/5

Is this solution Helpfull? Yes (35) | No (2)  

ARPIT SRIVASTAVA  (10 years ago)

Frnd manikandan...the 4 possible pairs will be like RRRW , RRWR , RWRR , WRRR
Hence 4* (3/5*2/4*1/3*1)
i.e., 4* 1/10

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gaurav kumar verma  (10 years ago)

soln:
probability of white balls always coming last is equal to probability of all red balls taken out at first
i,e. (3/5)*(2/4)*(1/3) = 1/10 Ans.

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ARPIT SRIVASTAVA  (10 years ago)

Its 2/5.
Keep 1 white ball fix at last
Then there will be 4 combinations
Hence 4* 1/10 = 2/5

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tausif  (10 years ago)

Actually here little bit mistake the question was
What is the Probability that one of the white balls always come at last??

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gaurav kumar verma  (10 years ago)

if any of the white ball can come last then
any of the three red balls and one of the white balls should be taken out first
i,e. (4/5)*(3/4)*(2/3)=2/5 Ans.

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mohit ranjan basak  (10 years ago)

the psssibility for favaourable events=R/W,R/W,R/W,R/W,W
so tatal arrangement=4!/3!=4
total num of events=5!/(3!*2!)=10
hence probabilty=4/10=2/5 ans....

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Dolly Singh  (9 years ago)

Here is my take.

We have 3 red and 2 White balls . We just want that in the 5th ball must be white.

We can visualize it as arrangement of 5 balls (of which 3 are red and 2 are white) in such a way that 5th ball must be white.

We can fix 1 white ball at 5th place and can arrange remaining balls (Total 4 = 3 Red + 1 White) as

4!/3! = This is our desired outcome

Probability = {Desired Outcomes}{Total Outcomes}

Desired Outcomes = 4

Total outcomes = 5!/3!2!= 10

Probability = 4/10 = 2/5

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Mani kandan  (10 years ago)

hai srivastava, how become 4* 1/10 pls explain

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tausif  (10 years ago)

There are two White balls in the bag so white also come at Any Position

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Sridhar Reddy  (10 years ago)

I think its 3/10.

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