# Where possible, give answeres as fractions. Where a decimal is given, it must be given corrrect to four decimal places. a) for this part of the question, state which rule of probability you use at...

*Where possible, give answeres as fractions. Where a decimal is given, it must be given corrrect to four decimal places.*

*a) for this part of the question, state which rule of probability you use at each stage in your calculations. For example P(E) = n(E)/N, or the multiplication rule, and so on.*

i) Three four-sided dice are thrown together. Each die is labelled with 1,2,3,4 on its four sides. Find the probability of obtaining a 'triple four' - that is, the probability that all three dice land with the face showining a four facing downwards.

ii)Find the probability of obtaining at least one 'triple four' in four tosses of the three dice.

### 1 Answer | Add Yours

Total number of possible outcome in this experiment N= `4^3=64`

Its sample space is S ={ (111),(112),(113),(114),............(444)}

(1) E='triple four' =(444)

P(E)=`(n(E))/N`

`=0.0156`

(2).At least one 'triple four' in four tosses of the three dice.It mean either 1 triple four ,2 triple four or 3 triple four or 4 triple four.

E= event getting triple four , F= not getting triple four

P(F)=1-P(E)= 1-1/64=63/63

Thus by Binomial distribution

probability of not getting triple four

`C(4,0)(1/64)^0(63/64)^4`

=`(63/64)^4`

Thus the probability of getting at least one triple four

= 1- Probability of not getting triple four

=`1- (63/64)^4`

`=.0611`