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

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mathewww's profile pic

Posted on

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.

pramodpandey's profile pic

Posted on

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

 

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