@SAHILVANSIL Please like answers in case they match with the original answers
@SAHILVANSIL said:7. The odds that A speaks the truth is 3:2 and the odds that B speaks the truth is 5:3. In what percentage of cases are they likely to contradict?
47.5 ??
@SAHILVANSIL said:7. The odds that A speaks the truth is 3:2 and the odds that B speaks the truth is 5:3. In what percentage of cases are they likely to contradict?
19/40 * 100 %
@SAHILVANSIL said:6.The odds against A solving a problem are 7:5 and the odds in favour of B solving it are 12:9. What is the probability that if both of them solve it, it will be solved.
3/4 ?
7. The odds that A speaks the truth is 3:2 and the odds that B speaks the truth is 5:3. In what percentage of cases are they likely to contradict?
@SAHILVANSIL said:7. The odds that A speaks the truth is 3:2 and the odds that B speaks the truth is 5:3. In what percentage of cases are they likely to contradict?
2/5*5/8+3/5*3/8=19/40 :neutral:
@vijay_chandola said:Deja vu Three unbiased dice I, II and III are rolled simultaneously. Assuming that the sum of the numbers on the dice is 15, what is the probability that the die I has shown a 4?
Possible values on the dice could be: (6,5,4), (6,6,3), (5,5,5) and the number of ways to arrange them are 6, 3, 1 respectively.
so totally 10 ways.
(4,5,6) and (4,6,5) are the two combinations
hence 2/10 = 1/5
8.What is the chance of drawing, without bias four aces successively from a pack of 52 cards.
@SAHILVANSIL said:7. The odds that A speaks the truth is 3:2 and the odds that B speaks the truth is 5:3. In what percentage of cases are they likely to contradict?
47.5 % ?
One more sitter :P A mutual fund allows new investors to open accounts or existing investors to deposit money in their accounts only at the beginning of the year. It credits interest at the end of the year on the balance (B) at the beginning of the year. If B is not more than | |||||
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@SAHILVANSIL said:8.What is the chance of drawing, without bias four aces successively from a pack of 52 cards.
1/13^4*4! ?
@vijay_chandola said:Deja vu Three unbiased dice I, II and III are rolled simultaneously. Assuming that the sum of the numbers on the dice is 15, what is the probability that the die I has shown a 4?
a+b+c=15
a'+b'+c'=12 ; ways=14c2
We have to omit cases in which any one of them is greater than 6
a+b+c=5; ways=3(7c2)
T'=14c2-3(7c2)=28
For favourable
b+c = 11
b'+c'=9 ways=10-6(omit>6)=8
4 can be in any 3 so favourable=12
P(E)=12/28 = 3/7?
a'+b'+c'=12 ; ways=14c2
We have to omit cases in which any one of them is greater than 6
a+b+c=5; ways=3(7c2)
T'=14c2-3(7c2)=28
For favourable
b+c = 11
b'+c'=9 ways=10-6(omit>6)=8
4 can be in any 3 so favourable=12
P(E)=12/28 = 3/7?
9. Assuming that ˝ of the population is vegetarian, 100 investigators take a sample of 10 individuals to whether they are vegetarians. How many investigators would you expect to report 3 or less vegetarians?
@Torque024 said:a+b+c=15a'+b'+c'=12 ; ways=14c2We have to omit cases in which any one of them is greater than 6a+b+c=5; ways=3(7c2)T'=14c2-3(7c2)=28For favourableb+c = 11b'+c'=9 ways=10-6(omit>6)=84 can be in any 3 so favourable=12P(E)=12/28 = 3/7?
:nono:
OA=1/5
@SAHILVANSIL said:8.What is the chance of drawing, without bias four aces successively from a pack of 52 cards.
4c4 / 52c4
@sowmyanarayanan said:Aarav, who lived on Earth, went to Mars and met Jadoo, Badoo, Kadoo and Ladoo there. All four of them, who were residents of Mars, had some chocolates with them. On the arrival of Aarav, they became so happy that all of them gave away all their chocolates to Aarav. The number of chocolates with Jadoo, Badoo, Kadoo and Ladoo were 15, 40, k01 and 122 respectively as per the number system prevalent on their planet. They also told Aarav that the number of chocolates they had was in an arithmetic progression in the number system used on their planet. Aarav was an expert in number systems. If Aarav counted the chocolates that he received, which of the following cannot represent the number of chocolates that he counted? 1)122 2)233 3)442 4)278 5)None of these
see sowmya i got it this way:
let the base of the system which they follow be n.
now accordingly, the numbers can be represented as:
n+5
4n,
kn^2+1,
n^2+2n+2
now adding all these, we'll get :
(k+1)^2+7n+8
now closely looking at the options (without which it would have been pretty complicated to solve, in my opinion), we see that 278 is represented as 2n^2+7n+8.
so if we put k=1 and n=1 in our equation we get the above number(i.e 278).
but since 1 cannot be the base of the system(otherwise other numbers wouldn't have been possible), we can safely conclude that 278 cannot be the sum in any case.
@vijay_chandola said:One more sitter A mutual fund allows new investors to open accounts or existing investors to deposit money in their accounts only at the beginning of the year. It credits interest at the end of the year on the balance (B) at the beginning of the year. If B is not more than र25000, the rate of interest is 10% p.a., otherwise it is 15%.If a man deposits र10000, र12000 and र15000 at the beginning of the first, second and third years, what is the worth of his deposits at the end of three years?
46345?
10. Find the chance of getting 3 successes in 5 trials when the cha.nce of getting a success in one trial is
2/3
2/3
11. In a bombing attack there is 50% chance that anyone bomb will strike the target. Two direct hits are required to destroy the target. How many bombs must be dropped to give a 99% chance or better of completely destroying the target.
@vijay_chandola said:let say in mars the base is 6.15=11 base 1040=24 base 10122=50 base 104th number will be 37 (for arithmetic progression)Hence, sum=11+24+37+50=112Now, 122 in base 10 in base 7 = 233122 in base 10 in base 5 = 442So, 278 cannot be the answer.
Vijay bhai... 6 kaise observe kiya? One more doubt .. if numbers in some base system are in A.P. , then will they also be in A.P. in base 10?