7c2 x 3 + 3 ?
@wontmissthstime said: Seven different objects must be divided among three people. In how many ways can this be done if one or two of them must get no objects?how can we solve this with gap method????
gap method to pata nhi ..it can be solved liked this
if one does not get any object
so selcting that person takes 3 ways..
now 7 DIFF objects can be distributed in 2^7 - 2
so 3(2^7 - 2)
two of them dnt get object so selecting dem in 3c2 = 3 ways..and 1 can get in 1 way all objects
so 3*3*(2^7 -2)
isme teeno ko mile vo thdi check krna he
@wontmissthstime said: @rkshtsuranawhy 2^7-2???
one object can go to 2 boys..2 object can go to 2 boyz..so total 2^7 and -2 for the case when all object go to one boy
@rkshtsurana said:gap method to pata nhi ..it can be solved liked thisif one does not get any objectso selcting that person takes 3 ways..now 7 DIFF objects can be distributed in 2^7 - 2so 3(2^7 - 2)two of them dnt get object so selecting dem in 3c2 = 3 ways..and 1 can get in 1 way all objectsso 3*3*(2^7 -2)isme teeno ko mile vo thdi check krna he
My only doubt is why you multiplied 3 with 3(2^7 - 3)?
Shouldn't they be added?
@rachit_28 said:My only doubt is why you multiplied 3 with 3(2^7 - 3)?Shouldn't they be added?
if u add a + a + a = 3a which is equal to 3*a
@Nirvana_MBA 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 A.P
My ans... make the total AP's takins first no. as 1... increasing the diffrence by 1
1 2 3
1 3 5
1 4 7
1 5 9
1 6 11
1 7 13
1 8 15
1 9 17
1 10 19
1 11 21
these are 10 similarly for 2 it will come 9 for 3 it will come 9..thn for 4 & 5 it will come 8 and so on....when u add these it will come 100 and total three number cn be selected as 21C3 therfore probability is 100/21C3 it is 10/133
@rkshtsurana said:if u add a + a + a = 3a which is equal to 3*a
Bhai 3+3*(2^7 -2) and 3*3*(2^7 -2) same kaha hue ?
@rkshtsurana said:are ha sry..ryt
Bhai main cheez to batai hi nhi aapne, add hona chahiye ya multiply ?
@shrikantmishr said:Please solve this que:
66 students of a class are covered under a Health Insurance Scheme for which premium charged is Rs. 5000 per student. Considering the types of coverage in the policy, the probability that a policy will become a claim policy is estimated to be 0.2. The Insurance Company offering the cover has collected the data related to treatment expenses of various diseases covered under the scheme. Based on this, the claim amount is likely to vary linearly between Rs. 10000 to Rs. 50000.
What is the probability that this scheme will result in a loss. The company keeps 20% of the premium amount collected for meeting overhead and other expenses and only 80% is available for claim settlement.
The Insurance Company wants to do such business only if the probability of loss is less than or equal to 20%.
Is the probability of loss in this case
@MikelArteta said: I think the insurance company should charge Rs. 10500 as premium to keep the probability of loss under 20 %. Originally, the probability of loss is 75%
gud 1 mikel
@shrikantmishr The earning out of each student is 5000-1000 = 4000. The probability of the claim is 0.2.
For zero loss, let the claimed amount be Rs.X . Therefore,
(0.2)X = 4000
X = 20000
So, if the amount of claim is Rs.20000, the Most probable case is 0% profit or loss.
The probability of loss = The probability that the claimed amount is more than 20,000
P = (50,000 - 20,000)/(50,000-10,000) = 0.75
BTW, what is the correct answer to this question ?
For zero loss, let the claimed amount be Rs.X . Therefore,
(0.2)X = 4000
X = 20000
So, if the amount of claim is Rs.20000, the Most probable case is 0% profit or loss.
The probability of loss = The probability that the claimed amount is more than 20,000
P = (50,000 - 20,000)/(50,000-10,000) = 0.75
BTW, what is the correct answer to this question ?
@priyagoyal2711 said:Can someone help me with this problem:In how many ways can you divide 10 identical chocolates between 3 people so that any one person does not get more than 5 chocolates.Thanks in advance.
is it 48?? will explain if it is correct
@priyagoyal2711 said: Can someone help me with this problem:In how many ways can you divide 10 identical chocolates between 3 people so that any one person does not get more than 5 chocolates.Thanks in advance.
Is the answer 21 ???
@priyagoyal2711 said: It will help, if you guys can tell me what methods you used to solve then, and then it can be figured out.
distributing 10 similar things to 3 people is same as solving a+b+c = 10 with condn that a,b,c>= 0 we get 12C2 = 66 ways , now this has cases when any one of these three people has more than or equal to 6 chocs so let that person be "a" so we have a>=6 means
a-6>=0 let a' = a-6 => a = a'+6 , now as a+b+c = 10 substituting a we get a'+6+b+c = 10 => a'+b+c = 4 which has 6C2 solns , we can select that one person in 3 ways so total invalid cases are 3 x 6C2 = 45 , so favorable cases are 66-45 = 21 hope m rite !!!
Regards,
Could anyone please explain this?? In how many way 6 different chocolates be distributed to 3 boys?
Would like to have a solution to the below question (if possible with explanation):
Mandeep has to create a password having 5 distinct characters using at least 2 digits (from 7 to 9) and at least 2 English vowels (from A, E, I, O and U). No character other than digits from 7 to 9 and vowels of English alphabets are allowed in that password. If the password starts with a digit, then it must end with an alphabet and if it starts with an alphabet, then it must end with a digit. Find the number of possible passwords that Mandeep can create:
Thanks in advance!!!!
(a) 1260 (b) 4480 (c) 3620 (d) 2880
Thanks in advance!!!!