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In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fth throw of the die is equal to:

 Apr 24, 2019
 #1
avatar+105654 
+3

It seems to me that this is going to be the same as rolling a dice 5 times and finding the probability that 4 is rolled any 2 times in a row.

call a 4 a success and any other a failure then   for any individual roll   P(S)=1/6     P(F)=5/6     

 

 

 

44444  Any 5 fours        1 way                                                   (1/6)^5

4444  any 4 fours           5 ways                                                 5*(1/6)^4*(5/6)

444      an 3 fours          5C3 ways = 10 ways                           10*(1/6)^3*(5/6)^2

except   4*4*4     no but all other 3 fours ok.    -1 way                (1/6)^3*(5/6)^2

No fours 1 way    (5/6)^5                                                               

P(4*4**)=P(4**4*)=P(4***4)=P(*4*4*)=P(*4**4)=P(**4*4) = (1/6)^2*(5/6)^3

 

 

\(\text {P(rolling 2 fours together in 5 rolls ) }\\ =\text{1-P(no fours)-P(1 four)-P(2 fours but not together) - P(3 fours but none together) } \\=1-(\frac{5}{6})^5- 5(\frac{1}{6})^1(\frac{5}{6})^4 - 6(\frac{1}{6})^2(\frac{5}{6})^3 - (\frac{1}{6})^3(\frac{5}{6})^2\)

 

1-(5/6)^5-5*(1/6)^1*(5/6)^4-6*(1/6)^2*(5/6)^3-(1/6)^3*(5/6)^2 = 0.0965792181069959   \(=\frac{751}{7776}\)

 

2 fours in the right places

44***

*44**

**44*

***44

P(2 fours in the right places) = 4*(1/6)^2(5/6)^3

 

OR

P(rolling 2 fours together)

=P(5 fours)+P(4 fours)+ P(3 fours) - P(4*4*4) +P(2 fours in the right places)

(1/6)^5+5*(1/6)^4*(5/6)^1+10*(1/6)^3*(5/6)^2-(1/6)^3*(5/6)^2+4*(1/6)^2(5/6)^3 = 0.0965792181069959    \(=\frac{751}{7776}\)

 

 

 

That is excellent.

I did it 2 different ways and got the same answer both times :)

 Apr 24, 2019
 #3
avatar
+1

thank you so much and sorry for the typo its not "FTH'' ITS ''FIFTH''

Guest Apr 26, 2019
 #2
avatar+28194 
+3

I interpret this question to mean the probability of the experiment ending exactly on the fifth throw. Melody has calculated the probability that it could end on any throw from the second to the fifth (I think!).

 

If my interpretation is correct, then the fourth and fifth throws must be 4s, and the third throw must be not4.   This means the possible combinations for the first two throws are (1) 4 followed by not4, (2) not4 followed by 4, (3) not4 followed by not4.

 

Putting all this in probability terms we have p = (1/62)(5/6)[ (1/6)(5/6) + (5/6)(1/6) + (5/6)(5/6) ]

or p = 175/7776 ≈ 0.0225

 

This is borne out by a MonteCarlo simulation (which also confirms Melody’s calculation using her interpretation of the question).

 Apr 24, 2019
edited by Alan  Apr 24, 2019

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