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The answer is 1/18
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is it 125/216 ?
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@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!
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problablity that bob wins is 5^3/6^4 i.e. 125/1296
correct?
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Sorry cooltwins, perhaps its best that you discuss your approach, that way others can also help build the solution!!
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No.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
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or 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
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I 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. 😀.
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but ishu in the question it clearly states that sue starts. 😐
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No, read the problem again. Ill IM you the hint 😀
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And 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
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Is answer 275/1296 ?
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Reya
Member •
Apr 8, 2011
@blunderboy: see the above post!!Answer is already given 😀
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@ 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 😔
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A 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?
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vibhor_one
A 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?
My guess is
200....😕
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how you solved it? please explain I am not getting the same answer
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vibhor_one
how you solved it? please explain I am not getting the same answer
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?
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yes answer is correct, but how you worked it out? I am not getting the same answer
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Well 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 😀
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thanks this solution really helped me to understand the concept
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The 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
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