### Re: BDMO 2017 National round Secondary 1

Posted:

**Tue Feb 20, 2018 7:11 am**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 thatnot 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 thanWWWpossible? 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.