# 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.

pramodpandey | Student

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`