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

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)
10 years agoHelpfull: Yes(4) No(1)
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
10 years agoHelpfull: Yes(3) No(18)
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
10 years agoHelpfull: Yes(2) No(3)
Use the binomial theorem:
ncr*p^r*q^r
4C2*(1/6)^2*(5/6)^2=25/216.
10 years agoHelpfull: Yes(1) No(1)
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.
10 years agoHelpfull: Yes(1) No(1)
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
10 years agoHelpfull: Yes(0) No(0)
4C2 (1/6)^2 (5/6)^2
= (6)(1/6)^2(5/6)^2
8 years agoHelpfull: Yes(0) No(0)

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