## Search found 100 matches

Sat Dec 10, 2011 12:25 pm
Forum: Higher Secondary Level
Topic: 3 problems of probability
Replies: 15
Views: 10539

### 3 problems of probability

1. if a student has passed all the 11 exams he attended before, what the probability of his passing the next exam? 2. if the student got 80%+ in his previous 11 exams, what is the probability of getting 80% marks in the next exam? 3. if the students would have passed in 10 out of 11 exams, what woul...
Wed Dec 07, 2011 12:36 pm
Topic: Jessore 2007 HS prob 11
Replies: 9
Views: 3417

### Re: Jessore 2007 HS prob 11

nafistiham wrote:এখানে তো $\angle DBA$ এর মানটা ধ্রুব থাকছে না । তাই না?
আমিও সেটাই বলছি :S
Tue Dec 06, 2011 9:23 pm
Topic: Jessore 2007 HS prob 11
Replies: 9
Views: 3417

### Re: Jessore 2007 HS prob 11

লাটেক্স ঠিক করছি। কিন্তু $90^{\circ}$ কেমনে হয়??
Tue Dec 06, 2011 7:46 pm
Topic: Jessore 2007 HS prob 11
Replies: 9
Views: 3417

### Re: Jessore 2007 HS prob 11

আমার মাথায় একটা সলুশন আসছে। দেখ তো:
Tue Dec 06, 2011 5:13 pm
Topic: Kustia 2009 HS prob 7
Replies: 6
Views: 2236

### Re: Kustia 2009 HS prob 7

At first I also thought that there were only 48. But look at these combos:: $ZXYZZXY$ and $XYZZXYZ$. In the same way they have 48 permutations. So the solution is $144$. hmm..... so there are total 3 combos, each having 48 permutations $XZYZXZY$ $ZYXZZYX$ $ZXYZZXT$ and the answer is 144 :D thanks L...
Tue Dec 06, 2011 4:11 pm
Topic: Gopalgang higher hisec 2011 10
Replies: 4
Views: 1964

### Re: Gopalgang higher hisec 2011 10

nafistiham wrote:but,amlan da
Euclid says,
the summations of two sides in a triangle is larger than the other side.

so, the rightmost $5$ combos are incorrect.
হায় হায়, এমন ভুল আমি কেমনে করলাম
thanx nafis
Tue Dec 06, 2011 2:34 am
Topic: Kustia 2009 HS prob 7
Replies: 6
Views: 2236

### Re: Kustia 2009 HS prob 7

nafistiham wrote: $XZYZXZY$
$or$
$YZXZYZX$

the number of $X,Y,Z$ are such $2,2,3$
so the number of combinations are
$2\cdot2\cdot3!=24$
nafis, as u have considered two situations(xzyzxzy & yzxzyzx) so you have to multiply $24$ with another $2$. so answer should be 48
Tue Dec 06, 2011 2:20 am
Topic: Jess-06-hisec-Q.10
Replies: 2
Views: 1743

### Re: Jess-06-hisec-Q.10

Tue Dec 06, 2011 1:45 am
Topic: Gopalgang higher hisec 2011 10
Replies: 4
Views: 1964

### Re: Gopalgang higher hisec 2011 10

but i think the answer is $8$ .
let $a, a, b$ are the sides of the triangle. so
$2a+b=20$
$a=\frac{20-b}{2}$
there are $8$ possible results of $(a,b)$ [$(9,2),(8,4),(7,6),(6,8),(5,10),(4,12),(3,14),(2,16),(1,18))$]( whereas a, b are positive integers.
Mon Dec 05, 2011 12:35 am
no two member of X is a square less or equal to 21^2. তবে এখানে যদি sqare কথাটার পরে একটা ,(কমা) বসে, তবে ব্যাপারটা ভিন্ন হয়ে যাবে। তখন উত্তর সম্ভবত $১১৮$ হবে। এক্ষেত্রে বিষয়টা এরকমঃ আমরা যেই ২২১টা জিনিস নিয়ে $X$ তৈরি করেছি, তার যেকোনো ২টির যোগফল পূর্ণবর্গ হবে না। এখন যেকোনো ২টির যোগফল যেহেতু \$>...