two students are selected at random one after the other without replacement, from a group of 10 students. there are 6 male students.
given that the second selected student is a male, what's the probability of the first selected student is male?
since they have said the second student is male,
the probability of picking a male student for the second time is
P(b2) = 6/10 * 5/9 + 4/10 * 6/9 = 54/90
cause the first student isn't mentioned I took as it could be either a girl or a boy
both selected students being boys, i took it as
P(b1 AND b2) = 6/10 *5/9 = 30/90
so the first boy being a boy given that second one is a boy
P(b1/b2) = 5/9
this is the answer i got. can someone please tell me whether this is the correct answer and the correct way of solving this problem?
Best Answer
Under the condition that one of the six males is the second selected student, there remain five males among the nine, all of whom have equal probability to have been the first selected student.
Clearly the conditional probability is $5/9$.
If you wish to use Bayes' Rule to confirm:
$$\mathsf P(M_1\mid M_2)=\dfrac{\mathsf P(M_1\cap M_2)}{\mathsf P(M_2)}=\dfrac{\tfrac 6{10}\tfrac {5}{9}}{\tfrac 6{10}\tfrac 59+\tfrac 4{10}\tfrac 69}=\dfrac 59$$