The probability of students appeared for CAT is 4/5 and for MAT is

[m.gayatri]

The probability of students appeared for CAT is 4/5 and for MAT is 2/5.then find out the prob. that student appeared for at least one test?

[mahul000]

reqd probablity:--- 22/25-th.

[gomesdhamu2006]

6/5

[anubhav1109]

22/25

[anubhav1109]

answer given by gomesdhamu is wrong

2 ways of solving this problem is:-

1.
required probabilit is =1-(probability of student not appearing any exm)
=1-1/5*3/5 =22/25

2.required probablility=4/5*3/5+1/5*2/5+4/5*2/5
=22/25

[jimish]

i m agree with Anubhav,,hey anubhav i got ur 1st one,,but not able to get 2nd one solution,,can u explain me ??

[anubhav1109]

hello...gimish.....the solution to 2nd answer is...

the required probability can be calculted as:-
P(appearing fr CAT)*P(not appearing fr MAT)+
P(not appearing fr CAT)*P(appearing fr MAT)+
P(appearing fr CAT)*P(appearing fr MAT)


hope u will understand now....

[ramc2008]

The answer cannot be determined - reason being there are students who have appeared in both exam and the probability of which is not specified.

Let's assume that there are 5 students. Since 4 appeared for CAT and 2 appeared for MAT, the possible combinations are:

3 CAT, 1 MAT, 1 Both
2 CAT, 2 Both, 1 None

In the first case, probability that student appeared for at least one test = 1

In the second case, probability that student appeared for at least one test = 4/5