Latest TCS Aptitude Question SOLUTION: In 4 throws of a pair of dice what is the chance of throwing a double twice?

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In 4 throws of a pair of dice what is the chance of throwing a double twice?

Asked In TCS nidhi gupta (9 years ago)

Solved swati gupta Read Solution (8)

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swati gupta  (9 years ago)

ans is 25/216
when a pair of dice is thrown,
probability of getting a double=6/36=1/6
probability of not getting a double will be=5/6
we have to get a double twice that means from 4 throws we have to select any two throws=4C2
probability=4c2 *(1/6)^2 * (5/6)^2= 25/216
(1/6)^2*(5/6)^2 means we are getting double 2 times and a non double 2 times

Is this solution Helpfull? Yes (35) | No (1)  

Souradip Nath  (9 years ago)

To get a double in a single throw = P(A) = 6/36 = 1/6
Compliment of P(A) = 5/6

So, ans is:

(1/6)*(1/6)*(5/6)*(5/6)

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

alok kumar sharma  (9 years ago)

ans 13/18
because firstly we will find when no double twice in any dices then we will
subtract from 1.

asuume 1 on the first dice then next three cant hold the 1 may be(2,3,4,5,6)
1/6*5/6*4/6*3/6 no duble twice in any one dices and this repeat 6th time so..

1/6*5/6*4/6*3/6)*6====5/18
1-5/18=13/18 ans

Is this solution Helpfull? Yes (3) | No (18)  

anshul sharma  (9 years ago)

probability of throwing a double in 1 throw is 1/6
hence in 4 throws probability of 2 doubles in 4 chances are 1/6*1/6*5/6*5/6=25/216

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

sruthi  (9 years ago)

Use the binomial theorem:
ncr*p^r*q^r
4C2*(1/6)^2*(5/6)^2=25/216.

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ranjan kumar  (9 years ago)

a pair of dice is thrown 4 times..so for a pair sample space would be 36(for 2 dices) now for throwing a double twice..it would be 6 as 11 22 33 44 55 66..
now these all 6 can come in single throw of 2 dices..so probablity will be 6/36 for 1 and for all it shud be 4 times..so probab will be 6/36 +6/36+ 6/36 + 6/36 =4/6 or 2/3.

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

Dharmendra Shukla  (9 years ago)

success=1/6=P
failure=5/6=Q
probability of getting double=nCr(P)^r(Q)^r=4c2 (1/6)^2(5/6)^2=25/216

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Debarghya dutta  (8 years ago)

4C2 (1/6)^2 (5/6)^2
= (6)(1/6)^2(5/6)^2

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