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Q. A speaks truth in 75% and B in 80% of the cases.In what percentage of cases they are likely to contradict each other in narrating the same incident?

Q.there are three events A,B,C one of which must happen and only one can happen at a time.
The odds are 8:3 againstA,5:2 against B,find the odds against C.

hey, do you have any idea which exercise these questions belong to .......these dont seem to be any near to what we have in text.

you may be having the answers , is the answer for the first one 35%........pls provide the answers

the second one is a total alien.

Are you thorough with the text book yet?? if not I suggest you complete them first... dont waste time with such new questions irrelevant to the text....

because the probablitity of an unknown question is less... .......hopefully

is the answer to the first question 45 % ??
if yes.. then mayb i can provide u wid the solution..
working on the second one !!!

yeah buddy the ist ans is 45%..........plz provide me with the sol
try to give the 2nd one too........i will be grateful to u

dealing with new que's is not a bad idea even if they are not from ur text book
u got to know more than to learn..................

u exactly don't know what type of que's board has set
it may be one of those two que's above............may be

well therez nothin really difficult abt it..

see...
when there is a contradiction.. either.. A speaks the truth and B lies.. or B speaks the truth and A lies..
thus..

percentage = (prob. of (A lieing when B speaks the Truth ) + ( prob. of B lieing when A speaks the truth) ) *100

now.. prob. of (A lieing when B speaks the Truth ) = P(c !d) = P(c intersection d)/P(d)
let P (c) = prob of A lieing
P(d) = prob of D speaking truth

now. P(c) = 0.25 P(d) = (O.8)
thus, P(c intersection d) = 0.25 * 0.8 = 0.2

hence,, prob. of (A lieing when B speaks the Truth ) = P(c !d) = 0.2/0.8 = 0.25

similaraly... prob. of B lieing when A speaks the truth) = 0.2* 0.75/ 0.75 = 0.2

thus...
percentage = (0.2+0.25)*100= 45


i'll try the other one !

Hey sachin,

these two events will be independent. A speaking whatever will not affect what B has to say......

The LP guide has this question.....they are independent.

For Q2 we have to find;

P(ABC')+ P(AB'C)+ p(A'BC)

iF we have the answer to the above, we can find the odds against C.

Pls confirm the question pkcit

yup kauser..
def the events are independent..tht's y i wrote this step...

now. P(c) = 0.25 P(d) = (O.8)
thus, P(c intersection d) = 0.25 * 0.8 = 0.2

which is only true for independent events !!