@chillfactor sir the answer 1.5 is correct......bt i'm unable to understand the soln. can u plz explain it
Expected weight = Summation of all {(weight gain)*(probability of gaining that much weight)}
Now, following cases needs to be considered -> he picked first capsule, probability is 1/3 and weight gained will be 1 kg
-> he picked second capsule, probability is 1/3 and weight gained will be 0 kg
-> he picked third capsule, prob is 1/3 and weight gain is 2 kg. But he has to pick one more capsule also, so 1/3 probability for first and second each and weight gain will be 1 and 0 respectively. So, overall 1/9 probability of gaining 3kg and 1/9 for 2kg.
-> he picked third capsule, then again he picked third capsule, so, now probability of gaining 5 kg is 1/27 (if he picks first capsule) and probability of gaining 4 kg will be 1/27 (if he picks second capsule)
and so on....
add all these cases and get the expected weight gain as 1.5
65% of students in a class have taken subject A, 80% B, 75% C & 85% D. At least what percentage of students have taken alll four subjects??Guys help me out with gud explanation
65% of students in a class have taken subject A, 80% B, 75% C & 85% D. At least what percentage of students have taken alll four subjects??
Guys help me out with gud explanation
If you add all the percentages,it would result in 305% ,this means that if all the four subjects are to be taken by particular student,then we have to subtract the percentage of students who have taken 3 subjects at most.
A point P inside an equilateral triangle, ABC is located at a distance of 3,4 and 5 from A, B , C respectively. What is the area of the triangle.@iLoveTorres try karo!!
Please find my solution in the attached pic. Area is 9+25 (root3)/4 sq. units
Alternatively there is a direct relation between the distances from vertex and the side length of an equilateral triangle given by 3(x^4+y^4+z^4+a^4)=(x^2+y^2+z^2+a^2)^2. So you may also plug in the values to find a^2 and hence the area of thr triangle.
Expected weight= Summation of all {(weight gain)*(probability of gaining that much weight)}Now, following cases needs to be considered-> he picked first capsule, probability is 1/3 and weight gained will be 1 kg-> he picked second capsule, probability is 1/3 and weight gained will be 0 kg-> he picked third capsule, prob is 1/3 and weight gain is 2 kg. But he has to pick one more capsule also, so 1/3 probability for first and second each and weight gain will be 1 and 0 respectively. So, overall 1/9 probability of gaining 3kg and 1/9 for 2kg.-> he picked third capsule, then again he picked third capsule, so, now probability of gaining 5 kg is 1/27 (if he picks first capsule) and probability of gaining 4 kg will be 1/27 (if he picks second capsule)and so on....add all these cases and get the expected weight gain as 1.5
I did this in a shorter way, but couldn't prove I was not making a mistake 😞 Fell asleep over it...
Bascially I said, let the required answer be x.
Now x consists of 3 cases, each with 1/3rd prob. Case 1 he gains 0 Case 2 he gains 1 Case 3 he gains 2 and starts the whole proces again so overall expectation which means an additional x. Hence here we can say he gains (2+x) So adding it up x = 1/3 [0 + 1 + (2+x)] = 1/3 (x+3) which gives x = 1.5
I am not getting the approach...can you please elaborate that why you have taken all 2,3,4,8 and left 1,5,6,7..???
It need not be this specific combo. Basically 1+8 = 9 2+7 = 9 3+6 = 9 4+5 = 9 So if we take 1 out of the 2 in each set above it is fine. 2, 4, 6, 8 will work, but so will 1, 2, 3, 4 or 1, 3, 5, 7, or even 1, 7, 6, 4...up to you. Basically I cannot have both types in a set present together as then we would get a total divisible by 9. regards scrabbler
A point P inside an equilateral triangle, ABC is located at a distance of 3,4 and 5 from A, B , C respectively. What is the area of the triangle.@iLoveTorres try karo!!
