Probability Problem
Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6 the
game is over. Sue rolls first, if she doesn't roll a 6, Bob rolls the die, if
he doesn't roll a 6, Sue rolls again. They continue taking turns until one of
them rolls a 6.
Bob rolls a 6 before Sue.
What is the probability Bob rolled the 6 on his second turn?
game is over. Sue rolls first, if she doesn't roll a 6, Bob rolls the die, if
he doesn't roll a 6, Sue rolls again. They continue taking turns until one of
them rolls a 6.
Bob rolls a 6 before Sue.
What is the probability Bob rolled the 6 on his second turn?
Replies
-
ISHAN TOPREThe answer is 1/18
-
vipandeepis it 125/216 ?
-
Prasad Ajinkya@ishutopre - no, could you explain the logic as to how you arrived at that no?
@vipandeep - no, although you seem to be thinking in the right direction! -
cooltwinsproblablity that bob wins is 5^3/6^4 i.e. 125/1296
correct? -
Prasad AjinkyaSorry cooltwins, perhaps its best that you discuss your approach, that way others can also help build the solution!!
-
cooltwinsNo.of way bob wins the game= 5*5*5*1
and total no. of ways=6*6*6*6
so probability = 5^3/6^4 -
cooltwinsor a simpler soln is:
if bob wins in 2nd round
probability=p(sue loses 1st round)*p(bob loses 1st round)*p(sue loses the 2nd round)*p(bob wins)
=5/6 * 5/6 * 5/6 * 1/6 -
ISHAN TOPREI will modify the answer
There are two probabilities in the way so that Bob will have second chance.
Sue, Bob, Sue, Bob
Also the probability of Bob having 6 out of 6*4 attempts is 1=1/24
So 1/24
So is the answer 1/24?
Modified again as CT suggested. 😀. -
cooltwinsbut ishu in the question it clearly states that sue starts. 😐
-
Prasad AjinkyaNo, read the problem again. Ill IM you the hint 😀
-
Prasad AjinkyaAnd here is the solution!
A: Sue wins
B: Bob wins
C: Bob wins in his second turn
D: Sue does NOT win in her first turn
We want to find the probability that Bob wins in his second turn, given that he wins in first place:
P(C|B) = P(C^B) / P(B)
Since B follows logically from C:
P(C|B) = P(C) / P(B)
The probability that the winner is determined in the fourth die toss (Bob’s second) can be easily calculated:
P(C) = (5/6) * (5/6) * (5/6) * (1/6) = 125/1296
Calculating P(B) is a little bit trickier, but not so much. First, we note that, with probability 1, either Sue or Bob must win the game at some point. (The probability that the game hasn’t before the nth die toss converges to zero as n grows. This doesn’t mean that it’s impossible that the game never ends.) And, since they can’t both win at the same time:
P(A) + P(B) = 1
Given that Sue doesn’t win in her first turn, Bob’s chances are exactly the same that they were for Sue before her first turn. So we have:
P(A) = P(B|D) = P(B^D)/P(D)
Since D follows logically from B:
P(A) = P(B|D) = P(B)/P(D) = P(B) * (6/5)
Replacing for P(A) in P(A) + P(B) = 1 gives:
P(B) * (6/5) + P(B) = 1
P(B) = 5/11
(Kudos to Nathanael Nerode for calculating the probability that Bob wins using geometric series!)
So we can calculate the original conditional probability:
P(C|B) = P(C) / P(B) = (125/1295) / (5/11) = 275/1296 -
Sachin JainIs answer 275/1296 ?
-
Reya@blunderboy: see the above post!!Answer is already given 😀
-
Sachin Jain@ Praveena
Hey thanks for pointing out but i just saw the problem and did not see the posts on second page.
I just saw the posts on first page and was not aware of post containing solution.
My bad 😔 -
vibhor_oneA candidate is required to answer 6 out of 10 questions, which are divided into two groups each containing 5 questions, and he is nott permited to attempt more than 4 from the group. In how many ways can he makw up his choice?
-
rahul69
My guess is 200....😕vibhor_oneA candidate is required to answer 6 out of 10 questions, which are divided into two groups each containing 5 questions, and he is nott permited to attempt more than 4 from the group. In how many ways can he makw up his choice? -
vibhor_onehow you solved it? please explain I am not getting the same answer
-
rahul69
I suppose u must be knowing the right answer, so is my answer correct? If it is correct then there is sense in explaining, otherwise not, and what is the answer u r getting?vibhor_onehow you solved it? please explain I am not getting the same answer -
vibhor_oneyes answer is correct, but how you worked it out? I am not getting the same answer
-
rahul69Well here is the solution: We have to select 6 questions from 10 questions divided in 2 groups of 5 and not more than 4 for each group,
So using combinations: {(5C4)*(5C2) + (5C3)*(5C3) + (5C2)*(5C4)}
= {5*10+10*10+10*5}
= {50+100+50}
= 200
Here, we got the answer.
Hope it helps 😀 -
vibhor_onethanks this solution really helped me to understand the concept
-
Saidu KonnehThe fact that Sue starts the game and that anyone playing a six ends the game, it implies that the total number of rolls is four(4), provided Bob plays six during his second roll.
So now: The probability of Bob playing a six in his second roll is equalled to the number of six's divided by the total number of rolls in the game: 1/4
You are reading an archived discussion.
Related Posts
first post,,
Top 5 Demotivating factors, in IT Company | Java | Entrepreneurship | Blogging | Forex
post your comments ,
Hi friends, I am preparing for GATE and want some advice on preparation. Some of my friends are going to Hyderabad for preparation. But what is the difference between self...
Are you still puzzled in enjoying MTS videos on iPod, iPhone, or other players? Then you must convert MTS to those players supporting formats. However,with high conversion speed and quality,...
hi, friends i am doing my final year in electrical and electronics engineering . i wanted to join in ISRO as a scientist in electronics feild . Which branch i...
Open Source (OS) software systems have taken over the whole wide world of the web by storm and have made websites more flexible and highly scalable. Of course, leading the...