Homework Help

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

user profile pic

andripol | eNoter

Posted April 27, 2013 at 9:48 PM via web

dislike 3 like

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

2 Answers | Add Yours

user profile pic

mathewww | High School Teacher | Honors

Posted May 3, 2013 at 6:07 PM (Answer #1)

dislike 1 like

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.

user profile pic

pramodpandey | College Teacher | Valedictorian

Posted May 25, 2013 at 6:32 AM (Answer #2)

dislike 0 like

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

 

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes