@ron123 said:Out of 21 tickets marked with numbers from 1 to 21 , three are drawn at random, find the probability that the three numbers on them are in AP.
10/133?
@ron123 said:Out of 21 tickets marked with numbers from 1 to 21 , three are drawn at random, find the probability that the three numbers on them are in AP.
c.d 1 - 19
c.d 2 - 17
c.d 3 - 15
...
total = 19+17+15+..+1 = 100
so 100/21c3
@ron123 said:Out of 21 tickets marked with numbers from 1 to 21 , three are drawn at random, find the probability that the three numbers on them are in AP.
If a, b, c are the three numbers in AP, then 2b = a + c.
Since the LHS is an even number, both a and c have to be either odd or even.
For every pair of even or odd values of (a,c), there will be one AP.
There are 11 odd numbers and 10 even numbers in the first 21 natural numbers.
Thus the answer will be (11C2 + 10C2)/21C3
Since the LHS is an even number, both a and c have to be either odd or even.
For every pair of even or odd values of (a,c), there will be one AP.
There are 11 odd numbers and 10 even numbers in the first 21 natural numbers.
Thus the answer will be (11C2 + 10C2)/21C3
@soumitrabengeri said:The combinations possible are1) 1 1 5--->7C1*6C1*3!/2!2) 1 2 4--->7C1*6C2*3!3) 1 3 3 --->7C1*6C3*3!/2!4) 1 0 6---->7C1*3!Add all the terms to get the answerPlease correct me if i am wrong
Seems right.... Can you explain me the calculation in Case 3.....
@soumitrabengeri said:If a, b, c are the three numbers in AP, then 2b = a + c.Since the LHS is an even number, both a and c have to be either odd or even.For every pair of even or odd values of (a,c), there will be one AP.There are 11 odd numbers and 10 even numbers in the first 21 natural numbers.Thus the answer will be (11C2 + 10C2)/21C3
When we selecting both even or both odd nos. Then those numbers could take either unit's place or tens place. So shouldn't we multiply with 2?
@ron123 said:Right ! Please share the approach ....
when 1 gets 1 object=7c1*3!*(6c6+6c4+6c3/2!)=42*(1+15+10)=42*26=1092
when 2 get 1 object=7c1*6c1*3!/2!=126
tot=1218
when 2 get 1 object=7c1*6c1*3!/2!=126
tot=1218
@ron123 said:How many minimum randomly arranged boys do we need to have such that if anyone of them is selected then there is at least 60% chance that he was born in a leap year.?(a) 3 (b) 4 (c) 5 (d) never possible.
1-(3/4)^4 = .683 > .6
4 boys
@ron123 said:Seems right.... Can you explain me the calculation in Case 3.....
1st person can get any of the 7 objects in 7C1 ways
2nd person gets 3 objects out of the remaining 6 in 6C3 ways
The remaining 3 objects go to the last person by default
These objects can then be arranged among the 3 persons in 3!/2! ways
So we get..7C1*6C3*3!/2!
@ron123 said:Correct.... Please share your approach....
If you start with 1 then APs can be 1,2,3.....
1,3,5.....
1,5,7.....
1,7,19.....
Total 10 ways
Similarly if you start with 2...then total ways 9
Similarly if you start with 3....then total ways 9
Similarly if you start with 4....then total ways 8
Similarly if you start with 5....then total ways 8
.
.
.
Similarly if you start with 19 ...then total ways 1
Total 10+9+9+8+8+...........1+1=100
Sample space=21c3
P=100/21c3=10/133
@ron123 said:How many minimum randomly arranged boys do we need to have such that if anyone of them is selected then there is at least 60% chance that he was born in a leap year.?(a) 3 (b) 4 (c) 5 (d) never possible.
(3/4)^n=>n=4?
@ron123 said:When we selecting both even or both odd nos. Then those numbers could take either unit's place or tens place. So shouldn't we multiply with 2?
Why the multiplication with 2?
Did not understand your doubt
@soumitrabengeri said:Why the multiplication with 2?Did not understand your doubt
Sorry.... Got it.... No need for multiplication with 2.... 
For all integers n ‰Ľ 1, a sequence of positive integers ai is defined by
.what is the minimum value of a1 + a2 ?
.what is the minimum value of a1 + a2 ?1) 213 2) 407 3)507 4)613
@wovfactorAPS said:answer seems to be right as shown by @chillfactor sir but(4n/5n)^3 has repititions,what say?
Can you show me how it has repetitions ? 😃
PS: Don't mind, but I think it's better to mention the whole thing in a single post that mention a part of the argument in a post :splat:. A lot of redundancy is already there, where "people just post the answer and not the approach, because it seems their self-confidence is a cause of concern" :/ . If one has doubts, he should not post just the answer. If one has made a mistake in the approach, then there is nothing better than that because the author of that post will remember the public humiliation and not make that mistake again, and isn't that the aim of this thread - "to learn" 😐 - this is a message/plea to all 😞
@soham2208 from now i will share my approach no matter how illogical it seems..yes i am here to learn 
@soham2208 said:Can you show me how it has repetitions ? PS: Don't mind, but I think it's better to mention the whole thing in a single post that mention a part of the argument in a post . A lot of redundancy is already there, where "people just post the answer and not the approach, because it seems their self-confidence is a cause of concern" :/ . If one has doubts, he should not post just the answer. If one has made a mistake in the approach, then there is nothing better than that because the author of that post will remember the public humiliation and not make that mistake again, and isn't that the aim of this thread - "to learn" - this is a message/plea to all
hmm. ok
say N=1 let numbers be 1 2 3 4 5
SO 5 numbers are there
4^3 will have sets like (2 3 4) (3 4 2).. would be counted
4^3 is basically an arrangement but here we are worried abt the product
say N=1 let numbers be 1 2 3 4 5
SO 5 numbers are there
4^3 will have sets like (2 3 4) (3 4 2).. would be counted
4^3 is basically an arrangement but here we are worried abt the product
what can be the maximum number of acute angles in an octagon in which all the interior angles are less than 180 degrees??
1
2
3
4
pls explain
@wovfactorAPS said:hmm. ok say N=1SO 5 numbers are there4^3 will have sets like (2 3 4) (3 4 2).. would be counted4^3 is basically an arrangement but here we are worried abt the product
See even if there is a change in order, it will count as a favorable/unfavorable cases in measuring the probability.
Here N is just a helping number. N in reality is a huge number. Let me explain my method in more detail
In this world, there are five types of integers
5k, 5k +1, 5k + 2, 5k + 3, 5k + 4
Now, let us the assume the total number of integers in this world is 5N , which is a huge number and N approaches infinity
Now, each type of integer will be equal to N (i.e. there are N integers of type 5k, 5k + 1 etc..)
So, consider the product of three numbers. It can be written in (5N)^3 ways
Also, unfavorable cases can be in (4N)^3 ways
So, unfavorable prob = (limit N-> inf) ( 4N/5N)^3 = 0.512
Favorable prob = 1 - 0.512 = 0.488
Now, (2,3,4) and (2,4,3) are two different unfavorable cases here and both will be counted in probability
@pavimai said:what can be the maximum number of acute angles in an octagon in which all the interior angles are less than 180 degrees??1 234pls explain
max of 3....
total angle in octagon =1080{ having 8 angles}
if we take max angle 179....suppose there are 4 sides having this angle...
4*(other angle) + 179*(4) =1080
other angles = 91 so not acut.....
it means at max there can be 3 acute angles...