A fair #6#-sided die is rolled up to #3# times as described below to give a score. If the score is #5# then what is the probability that it was obtained in #2# rolls?

Question

The die is rolled once.

If the first roll is #3, 4, 5, 6# then that is the final score.

If the first roll is #2#, then the die is rolled again and #2# added to give the final score.

If the first roll is #1#, then the die is rolled twice more, with the final score being the sum of all three rolls.

Nathan Axel · Accepted Answer

#2/15#

What a sweet inquiry!

To start, we really only need to take into account three possible outcomes.

#rarr# 3,4,5,6 is rolled. This is the score, Dice was rolled ONCE.

#rarr# 2 is rolled. Roll again and add the value. Dice is rolled TWICE

#rarr# 1 is rolled. Roll twice more and add on the values, The dice is rolled THREE times.

There are several ways of ending with a total of 5, but we are only concerned with an outcome of #5# from TWO rolls, so this can only come from the second option.

It is required that a 2 be rolled on the first roll and a 3 on the second.

#P(2,3) = 1/6 xx 1/6 =1/36#

This is the likelihood of receiving a 5 from two rolls. On the other hand, after giving it some thought, it appears that the question is more complex, as it asks, "IF a player receives a 5, what is the probability that it was from two rolls of the dice?" This isn't quite the same as the first part.

The player reportedly received a 5!, but how did he receive that score?

What is the number of ways to receive a 5?

From first outcome #P(5) = 1/6#

From second outcome. #P(5) = P(2,3) = 1/6xx1/6 =1/36#

Only the following methods are available for obtaining a 5 from the third outcome:

#1, 1, 3 " or " 1, 2, 2 " or " 1, 3, 1#

For each of these outcomes, the probability is #1/6xx1/6xx1/6 = 1/216#

So, for option 3, #P(5) = 3/216#

Therefore altogether, #P(5) = 1/6 + 1/36 + 3/216#

Utilize identical denominators:

#P(5) = 36/216 + 6/216 + 3/216 = 45/216#

Taking into account that out of 216 rolls, there exist 45 ways to obtain a 5, with only 6 arising from a mere 2 rolls.

Probability that two rolls will result in a score of five:

= #6/45 =2/15#

Eli Assouline · Answer

A different approach 2 of 2
Solved by counting

#2/15#

Remember that the 'given' part of a 'conditional probability' defines the sample space.

Using option 1 out of 2, we have #P(5_("a"))= 5/24 larr" of all possible scores"# #P(5_("b"))=1/36 larr" of all possible scores"# "Use any common denominator to determine the value of this variable." #P(5_("a"))= (5xx36)/(24xx36) = 180/846 larr" Sample space"# #P(5_("b"))=(1xx24)/(24xx36) = 24/846# So # P("5 due to path B")=24/180 larr" Just using counts"# So #color(blue)( P("5 due to path B")=2/15)#

Savannah Anderson · Answer

A different approach 1 of 2

#2/15#

A sketch usually helps visualise what is happening:

Key points;
2 roles of the die to reach a score #color(red)(=> " Path B")#

Final score is 5

#color(blue)("Diagram of all the paths that lead to a final score of 5")#

Thus the probability of any 5 occurring is #1/6+1/36+3/216 = 5/24#
Thus the probability of 5 from path B is #1/6xx1/6=1/36#

#color(green)("Condition 1")#
This is conditional probability in that it is categorically stated that the event: #color(green)("a 5 has occurred")#

#color(green)("Condition 2")#
What is the likelihood the #color(green)("5 came from path B")#

Let the probability of the 5 from path B be #P(5_"b")#
Let the probability of any 5 from all of them be #P(5_("a"))#

So we have probability event #P(5_"b")# given that #P(5_("a"))# has occurred.

This is written #P(5_("b")|5_("a"))#

But #P(5_("b")|5_("a")) = (P(5_("b")nn5_("a")))/(P(5_"a")) = (color(white)(.)1/36color(white)(.))/(5/24) -=1/36xx24/5 = 2/15#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Additional note")#

The 5 at the end of path B is shared ( a subset) between path B and the set of all the 5's. In my calculations this is written as #P(5_("b")nn5_("a"))# and has the value of occurrence of #1/36#

As it is shared it has the 'and' condition.

Probability #P(5_"b")" and "P(5_"a") ->P(5_("b") nn 5_("a"))#

Avery Alexander · Answer

#2/15#

This alternative perspective on the issue, in my opinion, helps to eliminate any uncertainty regarding conditional probabilities, etc.

Let's say we alter the game's rules without altering the likelihood of the results:

The impartial six-sided die is rolled precisely three times under the new regulations.

If the first roll was a #3#, #4#, #5# or #6# then that is the score for the turn.

If the first roll was a #2#, then the value of the second roll is added to give the score.

If the first roll was a #1#, then the value of all three rolls is added to give the score.

The odds of obtaining a given score are precisely the same as they would be if the die were not rolled again.

There are #6xx6xx6 = 216# equally probable possible sequences of values in #3# rolls.

Of these, #1xx6xx6 = color(blue)(36)# give a score #5# on the first roll, due to the first roll being #5#.

#1xx1xx6 = color(blue)(6)# give a score of #5# on the second roll, due to the first roll being #2# and the second #3#.

#1xx3xx1 = color(blue)(3)# give a score of #5# on the third roll, corresponding to the sequences: #1, 1, 3# or #1, 2, 2# or #1, 3, 1#.

So if the total score is #5# then the probability of it being one of the "two roll" scores is:

#6/(36+6+3) = 6/45 = 2/15#

Abigail Allen · Answer

undefined The probability of obtaining a score of 5 in 2 rolls of a fair 6-sided die is $ \frac{1}{6} \times \frac{5}{6} = \frac{5}{36} $.

A fair #6#-sided die is rolled up to #3# times as described below to give a score. If the score is #5# then what is the probability that it was obtained in #2# rolls?

The die is rolled once. If the first roll is #3, 4, 5, 6# then that is the final score. If the first roll is #2#, then the die is rolled again and #2# added to give the final score. If the first roll is #1#, then the die is rolled twice more, with the final score being the sum of all three rolls.