# Find the probability of obtaining at least one ‘triple four’ in fourtosses of the three dice.

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### 2 Answers

Answer to the first problem:

Probability of obtaining at least one ‘triple four’ in four

tosses of three dice will have to be found.

Clearly, the sample space of the experiment of a single toss of three dice is 6˄3 = 216. Out of these the event 'triple four' occurs only once. So, the probability of the desired event on one toss is 1/216.

On four such independent tosses, the probability is clearly 1*4/216 = 1/54.

**Hence probability of obtaining at least one ‘triple four’ in four tosses of three dice is 1/54.**

Probability of obtaining at least one ‘triple four’ in four tosses of three dice will have to be found.

Number of possible out comes in toss of three dice= 6^3=216

E= (444)

that is to say favourable out come =1

Thus

P(E)=1/216

Now in question at least one triple four in four tosse

E= 1 triplefour +2 triple four+3 triple four +4 trple foue

(444),(---),(---),(---)

(444),(444),(---)(---)

(444),(444),(444),(---)

(444),(444),(444),(444)

E'= no triple four in four tosses

P(E)+P(E')=1

But

P(E')=(1-1/216)^4=(215/216)^4

Thus

P(E=at least one triple four)=1-(215/216)^4

=0.01839