Last year,
$79\%$ of the students in a statistics class attended lectures regularly.
$93\%$ of students who attended lectures regularly passed the examination at the end of the module, but only
$19\%$ of the students who did not attend lectures regularly passed the examination.
a) A student has been selected at random from last year’s BS1 class. Calculate the probability that this student passed the examination.
My guess for A,
$P(A)$ – student passed and attend lectures
$P(B)$ – student passed and did not attend lecture
$P(A)\cdot0.79$ attend lectures + $P(B)\cdot0.21$ did not attend lectures = probability of this student passed the exam?
Correct me if I’m wrong thanks 🙂
b)Sam MacLeinn was one of the students who passed the examination. Calculate the probability that Sam attended lectures regularly.
Thanks in advance!!
Best Answer
Your thinking for a) is correct ... but your formula is not quite right. If you define $P(A)$ as the probability of a student attending and passing , and $P(B)$ the probability of a student not attending and passing, then the probability of passing is simply $P(A)+P(B)$.
That is, you don't further multiply $P(A)$ by $0.79$, but rather, you figure out $P(A)$ by multiplying the $0.97$ by $0.79$. That is:
$P(A)=0.97\cdot 0.79$
Likewise:
$P(B)=0.19\cdot 0.21$
For b): Note that $P(A)+P(B)$ gives you the percentage of students that passed the exam. So, what part of that is made up by students who attended regularly? That will be the probability of Sam attending regularly, given that Sam was one of those who passed. For example, if $P(A)$ turns out to be $.4$, i.e. $40$%, and $P(B)$ turns out to be $0.2$, i.e. $20$%, then out of all those students who passed the exam (which is $20+40=60$% of all stduents), the proportion that attended regularly is $\frac{0.4}{0.6}$