2

The Question

Consider a drawer with $6$ pens: $2$ blue, $2$ red, $1$ yellow and $1$ green. We choose two pens without replacement. What is the probability of extracting one blue pen second?

Answer given: $21/50$

enter image description here

I've tried many different methods, including finding the probability of picking a blue pen then a blue pen and then multiplying it by the probability of picking a blue pen first. Then adding that to the probability of picking a non-blue pen then a blue pen, then multiplying it by the probability of picking up a non-blue pen first. That method gives me $1/3$, and all other methods give me answers don't get close to the answer given.

Any help would be greatly appreciated!

Toby Mellor
  • 167
  • 3
    Not sure what "one blue pen second" means. If all it means is that the second draw is blue, then the answer is $\frac 26$. If it means that the second is blue AND the first is not-blue, then the answer is $\frac 46\times \frac 25$. I have no idea where the given answer comes from. – lulu Jan 16 '17 at 12:43
  • @lulu I've added a picture of the solution the lecturer gave us but I'm still unable to understand it, especially where the 1/30 came from. – Toby Mellor Jan 16 '17 at 12:48
  • 2
    That's because it is nonsense. The probability that the first draw is blue is $\frac 13$, obviously. The probability that the second draw is then also blue is $\frac 15$. Hence the probability $P(B_1\cap B_2)=\frac 1{15}$. What text is this? – lulu Jan 16 '17 at 12:52
  • @lulu Thanks for the clarification! The lecturer must be wrong. It's just weekly questions he sets – Toby Mellor Jan 16 '17 at 12:54
  • 2
    Just to stress: the probability that any given draw is blue is $\frac 26$. There is nothing special about being the first, second, or sixth draw. It's fine if the author wants to carefully go through the law of total probability to confirm this...that's a good exercise. Just do it carefully. – lulu Jan 16 '17 at 12:55
  • 1
    Oh, ok. Anybody can make a mistake at the board. – lulu Jan 16 '17 at 12:55
  • 2
    As an alternate way of computing $P(B_1\cap B_2)$: suppose the pens are numbered $b_1,b_2,r_1,r_2,y,g$. Then there are $6!$ equally probable ways of ordering them. There are $2\times 4!$ ways of ordering them assuming the two blues come first (in some order). Thus the answer is $\frac {2\times 4!}{6!}=\frac 2{6\times 5}=\frac 1{15}$. – lulu Jan 16 '17 at 12:58

1 Answers1

2

Your law of total probability is wrong. The right formula is

$$\boxed{P(B_2)=P(B_2|B_1)\cdot P(B_1)+P(B_2|B_1^C)\cdot P(B_1^C)}$$

That´s a big mistake from your lecturer.

$=\frac15\cdot \frac{2}6+\frac{2}{5}\cdot \frac46=\frac2{30}+\frac8{30}=\frac{10}{30}=\frac13$

Interpretation of $P(B_2|B_1)$:

One blue pen is drawn first. $5$ pens are remaining. One of them is blue. Thus the probability that the second drawn pen is blue is $\frac15$, given that the first drawn pen was blue.

callculus42
  • 30,550