probability

Here is how I did it:
First, I figured that since A speaks truth 60% of the time and B speaks truth 70% of the time and from the fact that they both tell the same lie...I figured that there are only two possible ways for a contridiction to occur.

First way: A tells truth and B lies
Second way: B tells truth and A lies

First way solve: (6/10)*(3/10)..this gives you 9/50 and the 6/10 came from A telling the truth and the 3/10 came from probability that B will lie. You multiply these together because it's a proability question.

Second way solve: (7/10)*(4/10).. this gives you 14/50 and the 7/10 came from B telling the truth and 4/10 came from probability that A will lie.

Third step is to combine 9/50 with 14/50. this gives you 23/50 or 46/100 = 46%

Answer is C

mitaliisrani wrote:A speaks truth in 60% of the cases while B speaks truth in 70% of the cases. When they lie they say the same lie. In what percent cases are they likely to contradict each other?
A) 30%
B) 40%
C)46%
D) 54%
E) 80%

ANs: OA is C

or A speaks the truth and B speaks lies or A speaks lies and B speaks the truth
1. A speaks the truth and B speaks lies = 60%*30%=18%
2. A speaks lies and B speaks the truth = 40*70/10000=28%
total= 18+28=46
C is the answer

mitaliisrani wrote:A speaks truth in 60% of the cases while B speaks truth in 70% of the cases. When they lie they say the same lie. In what percent cases are they likely to contradict each other?
A) 30%
B) 40%
C)46%
D) 54%
E) 80%

ANs: OA is C

Hi fieinds! what may be the answer when the Q is as follows:

A speaks truth in 60% of the cases while B speaks truth in 70% of the cases. When they lie they say the same lie. In what percent cases are they likely to contradict each other?

thanks