Latest Tech Mahindra Aptitude Question SOLUTION: Three faces of a fair die are Yellow, two faces red and one blue.The die is tossed three times. The probability that the colours, yellow, red and blue appear in the first, second a

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(#M40010745) TECH MAHINDRA Numerical Ability question arithmetic Keep an EYE

Three faces of a fair die are Yellow, two faces red and one blue.The die is tossed three
times. The probability that the colours, yellow, red and blue appear in the first, second and
third tosses respectively is

A)1/36 B)1/18 C)1/32 D)1/37 E)None of these

Asked In Tech Mahindra SATHISH T.N (11 years ago)

Unsolved Read Solution (6)

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PARAG KAMRA  (11 years ago)

Probability of getting yellow color = P(Y) = 3/6 = 1/2
Probability of getting red color = P(R) = 2/6 = 1/3
Probability of getting blue color = P(B) = 1/6

So, the probability that the colors yellow, red and blue appear in first, second and third toss respectively= P(Y ∩ R ∩ B) = P(Y) x P(R) x P(B) = 1/2 x 1/3 x 1/6 = 1/36(ans)

Is this solution Helpfull? Yes (37) | No (6)  

SDeep Bhattacharya  (11 years ago)

Since, the die is tossed thrice ,so total number of possible outcomes : 6^3 = 216.
Now,
In the 1st toss,we can get a yellow face in 3 ways.
For each of these 3 ways,
In the 2nd toss,we can get a red face in 2 ways.
So, we can get yellow in the 1st and red in the 2nd tosses in 3*2=6 ways.
And for each of these 6 ways,
In the 3rd toss,we can get a blue face in 1 way.
So, total no. of favourable cases w.r.t our desired event = 3*2*1=6.

So, by the classical definition of the probability ,
the required probability = (6/216)= (1/36)

So, the correct Option is : A)(1/36)

Is this solution Helpfull? Yes (19) | No (4)  

sahasra  (11 years ago)

given that 3 yellow faces,2 blue faces
dice are thrown 3 times
prob=(3c1*2c1*1c1)/(6*6*6)
ans:1/36

Is this solution Helpfull? Yes (5) | No (1)  

Swaraj Behera  (11 years ago)

Probability of getting yellow color = P(Y) = 3/6 = 1/2
Probability of getting red color = P(R) = 2/6 = 1/3
Probability of getting blue color = P(B) = 1/6

So, the probability that the colors yellow, red and blue appear in first, second and third toss respectively= P(Y ∩ R ∩ B) = P(Y) x P(R) x P(B) = 1/2 x 1/3 x 1/6 = 1/36

Is this solution Helpfull? Yes (2) | No   

PARAG KAMRA  (11 years ago)

Probability of getting yellow color = P(Y) = 3/6 = 1/2
Probability of getting red color = P(R) = 2/6 = 1/3
Probability of getting blue color = P(B) = 1/6

So, the probability that the colors yellow, red and blue appear in first, second and third toss respectively= P(Y ∩ R ∩ B) = P(Y) x P(R) x P(B) = 1/2 x 1/3 x 1/6 = 1/32(ans)

Is this solution Helpfull? Yes (1) | No (17)  

shamlik husna anees  (7 years ago)

die is tossed 3 times p(n)=216
=3x2x1/216
=1/36

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