Your solution to both b and c is wrong.SMMamun wrote: ↑Thu Feb 08, 2018 1:35 amThe key to understanding this problem, as Ahmed pointed out, is that not all cases are equally likely to happen. The probabilities for the series to end in exactly 3 matches and to end in exactly 4 (or 5) matches are not equal, so the 20 cases listed above cannot be treated with equal weights. In general, extra care must be exercised when the no. of elements in probabilistic events changes.
Listing of the 20 cases by Siam and Tasnood are appreciable and are worthy of Secondary level at which students should elaborate and analyze problems as detailed as possible even if that means spending additional time. Now let's see how we can approach the problem.
(a) This part is easier: simple, independent probability.
Bangladesh must win the 1st match. What is the probability of this? It's $\frac{1}{2}$. End of the story? No, not yet!
Bangladesh must win the 2nd match. Probability of winning the 2nd match is again $\frac{1}{2}$. What is the total probability of winning the first 2 matches? $\frac{1}{2} × \frac{1}{2}$.
What is the probability of WWW then? It's $\frac{1}{2} × \frac{1}{2} × \frac{1}{2} = \frac{1}{8}$
Is any scenario / case other than WWW possible? NO. So the ultimate probability is $\frac{1}{8}$.
(b) Bangladesh must with the 4th match and lose exactly 1 in the first 3 matches. You have already listed all the three combinations:
LWWW
WLWW
WWLW
What is the probability in the first case? $\frac{1}{2} × \frac{1}{2} × \frac{1}{2} × \frac{1}{2} $= $\frac{1}{16}$
So the total probability in all three cases = $3 × \frac{1}{16} = \frac{3}{16}$
If Bangladesh and India have different probabilities, say 70% for Bangladesh winning and 30% for India winning, you would just apply those percentages appropriately in places of W and L.
(c) If the series does not end in the first 3 or 4 games, it must end in the 5th.
Probability of ending in 3 matches = Probability that Bangladesh wins or (i.e. +) Probability that India wins = $\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$
Probability of ending in 4 matches = Probability that Bangladesh wins or (i.e. +) Probability that India wins = $\frac{3}{16} + \frac{3}{16} = \frac{3}{8}$
Probability of ending in 5 matches = $1  \frac{1}{4}  \frac{3}{8} = \frac{3}{8} = 0.375$
Another way to solve part (c): The series must have been tie after 4 matches: each team has won exactly two matches and lost exactly two matches, and the series goes to 5th match, 100% for sure. That can happen only in 6 possible cases ($4C2$ or $\frac{4!}{2!2!}$) if we consider 4 matches because in that case we do not need to know the outcome of the 5th match :
LLWW, WWLL
LWLW, WLWL
LWWL, WLLW
And the probability is $(6 × \frac{1}{16})$ or $0.375$.
Even another way to solve (c). There are 12 cases if we consider the outcome of the 5th match also.
WWLLW
WLWLW
WLLWW
LWWLW
LWLWW
LLWWW
WWLLL
WLWLL
WLLWL
LWWLL
LWLWL
LLWWL
And the probability is $(12 × \frac{1}{32})$ or $0.375$.
Hope it's clear.
BDMO 2017 National round Secondary 1
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 1
Frankly, my dear, I don't give a damn.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BDMO 2017 National round Secondary 1
Let,
The winning probablity of Bangladesh =$P(B)$
The winning probablity of India=$P(I)$
($a$)$\dfrac 18$
Bangladesh has played $3$ matches and won $3$ of them.It can be happened in $\dbinom 33 =1$ ways.Bangladesh has won $3$ matches.
So,$P(B)× P(B)× P(B)=\dfrac 12 ×\dfrac 12×\dfrac 12=\dfrac 18$
So,the desired probablity$=1×\dfrac 18=\dfrac 18$
($b$)$\dfrac {3}{16}$
Bangladesh has played $3$ matches and won $2$ of them.It can be happened in $\dbinom 32=3$ ways.Bangladesh has won $2$ matches and India has won $1$ match.
So,$P(B)× P(B)×P(I)=\dfrac 12×\dfrac 12×\dfrac 12=\dfrac 18$
As,$P(B)=\dfrac 12$
The desired probablity$=3×\dfrac 18×\dfrac 12=\dfrac {3}{16}$
($c$)$\dfrac{3}{16}$
Bangladesh has played $4$ matches and won $2$ of them.It can be happened in $\dbinom 42 =6$ways.Bangladesh has won $2$ matches and India has won $2$ matches.
So,$P(B)× P(B)× P(I)× P(I)=\dfrac 12×\dfrac 12×\dfrac 12×\dfrac 12=\dfrac {1}{16}$
As,$P(B)=\dfrac 12$
The desired probablity$=6×\dfrac {1}{16}×\dfrac 12=\dfrac {3}{16}$
The winning probablity of Bangladesh =$P(B)$
The winning probablity of India=$P(I)$
($a$)$\dfrac 18$
Bangladesh has played $3$ matches and won $3$ of them.It can be happened in $\dbinom 33 =1$ ways.Bangladesh has won $3$ matches.
So,$P(B)× P(B)× P(B)=\dfrac 12 ×\dfrac 12×\dfrac 12=\dfrac 18$
So,the desired probablity$=1×\dfrac 18=\dfrac 18$
($b$)$\dfrac {3}{16}$
Bangladesh has played $3$ matches and won $2$ of them.It can be happened in $\dbinom 32=3$ ways.Bangladesh has won $2$ matches and India has won $1$ match.
So,$P(B)× P(B)×P(I)=\dfrac 12×\dfrac 12×\dfrac 12=\dfrac 18$
As,$P(B)=\dfrac 12$
The desired probablity$=3×\dfrac 18×\dfrac 12=\dfrac {3}{16}$
($c$)$\dfrac{3}{16}$
Bangladesh has played $4$ matches and won $2$ of them.It can be happened in $\dbinom 42 =6$ways.Bangladesh has won $2$ matches and India has won $2$ matches.
So,$P(B)× P(B)× P(I)× P(I)=\dfrac 12×\dfrac 12×\dfrac 12×\dfrac 12=\dfrac {1}{16}$
As,$P(B)=\dfrac 12$
The desired probablity$=6×\dfrac {1}{16}×\dfrac 12=\dfrac {3}{16}$
Re: BDMO 2017 National round Secondary 1
The approaches are quite hard, mate, I must say,but whatever you do ,in whatever way the game goes ,I found that quite annoyingly the answer is
(a) 12.5%
(b) 100/16 %
(c) 100/3 %
answers given by Absur Khan Siam are right when the condition is applied that the players have to play 5 matches no matter what or who wins after 3 or rounds.
(a) 12.5%
(b) 100/16 %
(c) 100/3 %
answers given by Absur Khan Siam are right when the condition is applied that the players have to play 5 matches no matter what or who wins after 3 or rounds.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 1
This is the correct solution.ahmedittihad wrote: ↑Tue Feb 06, 2018 1:52 pm(a) Bangladesh wins the series in $3$ matches iff they win all $3$ of the games. And the probability of that is $\dfrac{1}{8}$.
(b) Out of the $16$ cases of a $4$ length binary string, Only 3 of them satisfy (WWLW, WLWW,LWWW). So the probability here is $\dfrac{3}{16}$.
(b) Probability of winning the series for Bangladesh is $\dfrac{1}{2}$. We can find the answer to this question by subtracting the answers of a and b from $\dfrac{1}{2}$. And, $\dfrac{1}{2}\dfrac{3}{16}\dfrac{1}{8}=\dfrac{3}{16}$
Frankly, my dear, I don't give a damn.
Re: BDMO 2017 National round Secondary 1
@ahmedittihad, I would rather say you are wrong. I have provided quite detailed clarification with easytofollow explanations. You have just drawn a conclusion without rebutting my arguments.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 1
Yah well knock me on facebook if you want. I'll explain in detail. Sorry, I can't write an in detailed post about what you're doing wrong.
Frankly, my dear, I don't give a damn.
Re: BDMO 2017 National round Secondary 1
@ahmedittihad,
I know where you have erred. In (c), you are calculating the probability of the series ending in 5 matches with the condition that Bangladesh must win the series. Nowhere in (c) is it mentioned that Bangladesh must win the series.
About your invitation to a nonrelevant forum, I have passed long, long ago those school days when precocious students resort to rambling discourses to avoid logical argument in a relevant forum. Let’s keep our BdMO forum relevant and lively, not digress from here. PERIOD.
I know where you have erred. In (c), you are calculating the probability of the series ending in 5 matches with the condition that Bangladesh must win the series. Nowhere in (c) is it mentioned that Bangladesh must win the series.
About your invitation to a nonrelevant forum, I have passed long, long ago those school days when precocious students resort to rambling discourses to avoid logical argument in a relevant forum. Let’s keep our BdMO forum relevant and lively, not digress from here. PERIOD.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 1
SMMamun math pare na chi chi chi
Frankly, my dear, I don't give a damn.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 1
And I'm better than you.
Frankly, my dear, I don't give a damn.

 Posts: 29
 Joined: Thu Jun 02, 2016 6:14 pm
Re: BDMO 2017 National round Secondary 1
Your solution is equally likely because its symmetrical.Absur Khan Siam wrote: ↑Sun Feb 04, 2018 12:34 pmMy solution may be wrong.But the $20$ cases yield to a result that both team have a probability of $\frac{1}{2}$ to win the series.
Means for a case WWWLL you have another case LLLWW.So its surely $1/2$ for both team.
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]