What is the probability that he picks up a blue pen? A boy has color weakness, so he is not good at distinguishing blue and red. There are $60$ blue pens and $40$ red pens randomly in a box. Given that he picks up a blue pen, there is a $60$% chance that he thinks it is a blue pen and a $40$% chance that he thinks it is a red pen. Given that he picks up a red pen, there is a $80$% chance that he thinks it is a red pen and a $20$% chance that he thinks it is a blue pen.
(a) What is that probability that he picks up a blue pen and recognizes it as a blue pen?
(b) What is the probability that he chooses a pen and thinks it is blue?
(c) Given that he thinks he chooses a blue pen, what is the probability that he actually chose a blue pen?
My work:
(a) Would it be $(1/2)*(6/10)$ = $3/10$
(b) $(1/2)*(3/5) + (1/2)(1/5) = 0.4$
(c) $50$%? 
I am not exactly sure how to approach these problems.
 A: You are almost correct but instead of using 1/2 each time, note that the probability of choosing a blue pen(correctly) at random isn't 1/2 it's $60/100$
So the answer to the first part is $\frac{9}{25}$. Now can you continue?
A: Let $B,R$ be the events that he picks a blue and red pen respectively. Let $\mathcal B,\mathcal R$ be the events that he thinks the pen is blue and red respectively. Then
(a) We have
$$P(B,\mathcal B) = P(\mathcal B|B)P(B) = .6(.6) =\frac{9}{25}= .36$$
(b) I will rephrase this, in other words, what is the probability that he thinks a pen is blue? Then he either picks blue and thinks blue or he picks red and thinks blue. Since these events are mutually exclusive, we have
\begin{align*}
P(\mathcal B) &= P(B,\mathcal B)+P(R,\mathcal B)\\
&= P(\mathcal B|B)P(B)+P(\mathcal B|R)P(R)\\
&=.6(.6)+.2(.4)\\
&=\frac{11}{25}\\
&=0.44
\end{align*}
(c) We have
$$P(B|\mathcal B) = \frac{P(B,\mathcal B)}{P(\mathcal B)} = \frac{9/25}{11/25} =\frac{9}{11} = 0.8181818$$
where in the third step I used parts (a) and (b).
A: Let $S$ be the event of selecting a blue pen, and $T$ be the event of thinking the pen is blue.   Their complements are doing the same for red pen.
Since there are 60 blue pens and 40 red pens we know the probability of selection at random.   We have also been given the probabilities of assessment after selection at random.
We have been given: $\mathsf P(S) = 0.60, \\ \mathsf P(S^\complement)=0.40, \\ \mathsf P(T\mid S) = 0.60, \\ \mathsf P(T^\complement\mid S)=0.40, \\ \mathsf P(T^\complement\mid S^\complement)=0.80 \\ \mathsf P(T\mid S^\complement)=0.20$


(a) What is that probability that he picks up a blue pen and recognizes it as a blue pen?

(a) Would it be $(1/2)*(6/10)$ = $3/10$

Right idea, wrong numbers.
$$\mathsf P(S\wedge T) = \mathsf P(S)~\mathsf P(T\mid S) = 0.36$$


(b) What is the probability that he chooses a pen and thinks it is blue?

(b) $(1/2)*(3/5) + (1/2)(1/5) = 0.4$

So again.
$$\mathsf P(T) = \mathsf P(S)~\mathsf P(T\mid S)+\mathsf P(S^\complement)~\mathsf P(T\mid S^\complement) = 0.44$$


(c) Given that he thinks he chooses a blue pen, what is the probability that he actually chose a blue pen?

(c) $50\%$? 

Did you try to use Bayes' Rule?

 $$\mathsf P(S\mid T) = \dfrac{\mathsf P(S)~\mathsf P(T\mid S)}{\mathsf P(T)}= \frac 9{11}$$

