Latest Elitmus Aptitude Question SOLUTION: Three dices rolled . what is the probability of getting atleast one six. Option a) 1/6*5/6*5/6 b) 3*1/6*5/6*5/6 c) 5/6*5/6*5/6 d) none

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(#M40033159) ELITMUS Numerical Ability question Elitmus 23rd feb14 Keep an EYE

Three dices rolled . what is the probability of getting atleast one six.
Option
a) 1/6*5/6*5/6
b) 3*1/6*5/6*5/6
c) 5/6*5/6*5/6
d) none

Asked In Elitmus Vivek Jain (10 years ago)

Unsolved Read Solution (9)

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Vivek Jain  (10 years ago)

probability of getting atleast one siz is 1-Probability if no six.
hence 1- 5/6*5/6*5/6

Is this solution Helpfull? Yes (30) | No   

sid  (10 years ago)

none... as prob. of getting 1 six is 1/6*5/6*5/6 two six is 1/6*1/6*5/6..three six is 1/6*1/6*1/6 and answer will be their sum. so none(ans)

Is this solution Helpfull? Yes (9) | No (1)  

Naveen Agarwal  (10 years ago)

none.....answer is 1-(5/6+5/6+5/6)=91/216 so "d" option

Is this solution Helpfull? Yes (3) | No (2)  

Abhijeet Vishnu Agale  (10 years ago)

And=*, OR=+ (1/6&5/6&5/6) OR (1/6&1/6&5/6) OR (1/6&1/6&1/6)= 31/216.
So,Ans is d.

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HARISHANKAR YADAV  (9 years ago)

AS Bionomial Distribution Formula:P(x=r)=nCr*p^r*(1-p)^r
p=sucess in A single trial.
q=Unsucess in a single trial(q=1-p)
ATQ:probablity of getting Six in singal trial p=1/6 nd q=5/6
given no dice rolled=3(n)
X=event of getting six.
nw we have calculate P(X>=1)=1-P(x=0)
P(X>=1)=1-(3C0*(1/6)^0*(5/6)^(3-0)).
P(X>=1)=1-((5/6)*(5/6)*(5/6))

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Mani kandan  (10 years ago)

ans is c.
Multiply the chances of NOT getting a six(5/6) on the first die by the chances of NOT getting a six on the second die (5/6) by the chances of NOT getting six on the third die(5/6), and the subtract the whole thing from 1.

1 - (5/6 x 5/6 x 5/6) = 0.5787

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krishna saxena  (10 years ago)

answer is "b"
question demands for atleast one six.
Prob of getting 6 in 1st dice X prob of not gettng six in 2nd dice X prob of not gettng six in 3rd dice
similarly,
not getting in 1st dice X getting in 2nd dice X not getting in 3rd dice
not getting in 1st X not gettng in 2nd X getting in 3rd

thereforem,
3(1/6*5/6*5/6) is correct

Is this solution Helpfull? Yes (1) | No (4)  

HARISHANKAR YADAV  (9 years ago)

SOrry above formula has missing some term
correct one is P(x=r)=nCr*p^r*(1-p)^(n-r)

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rohit kumar  (8 years ago)

(6+other+other)+(6+6+other)+(6+6+6)
(1/6*5/6*5/6)3!/2!+(1/6*1/6*5/6)3!/2!+(1/6*1/6*1/6)3!/3!
75/216+25/216+1/216
91/216

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