There are 3 drawers in a dresser`, and you are equally likely to pick any of the three. In drawer 1, there are 2 black socks and 3 red socks. In drawer 2, there are 3 black socks and 2 red socks. In drawer 3, there are 3 black socks and 3 red socks. Once you have randomly selected a drawer, you randomly pull out a sock of that drawer.

  1. If you were to randomly choose a drawer and the draw two socks from that drawer, what is the probability that you get a pair (red or black)?

Where is my mistake:

P(R OR B|D1)= 2/5 * 3/4 = 3/10 
P(R OR B|D2)= 3/5 * 2/4 = 3/10
P(R OR B|D3)= 1/2 * 1/2 = 1/4

P(R OR B)=P(R OR B|D1) P(D1) + P(R OR B|D2) P(D2) +P(R OR B|D3) P(D3) 

The correct answer is 0.4, but if I use my work I can not reach that value. Can you help me to know where is my mistake? Thanks, comunity.

  • $\begingroup$ The probabilty of P(R OR B|D1) = P(R|D1) + P(B|D1) = 2/5*1/4 + 3/5*2/4 = 1/10 + 3/10 = 2/5. etc. $\endgroup$
    – fleablood
    Nov 29 '15 at 3:00
  • $\begingroup$ Amusingly, the probability of a pair from the third drawer is also $2/5$, so calculation at the end is unnecessary. $\endgroup$ Nov 29 '15 at 3:05

For example you calculated $$P(R\text{ or }B|D_1)$$ as $3/10$. That is wrong. It should be $$P(R\text{ or }B|D_1) = \frac{1}{10}+\frac{3}{10} = \frac{4}{10}.$$ There rest are probably similar mistakes. The very last line you gave is correct.

  • $\begingroup$ If I calculate P(R OR B|Di) for each drawer I have P(R OR B|Di)= P(R OR B and Di)= 2/5 =0.4 for each draw, therefore any draw I choose have the same probability. Is that could be right? Futhermore, P(R OR B|Di)= P(R OR B AND Di) , mulitiplication rule. no? Thank, for your help. $\endgroup$
    – Electro82
    Nov 29 '15 at 3:14
  • $\begingroup$ I made a silly mistake! Hahaha. Sometimes I go on autopilot. Yeah, that's right. However, you wrote "P(R OR B|Di)= P(R OR B AND Di)" That's not true. It is true that given a draw, the probability of a pair is .4 for each one. $\endgroup$
    – Em.
    Nov 29 '15 at 3:46
  • 1
    $\begingroup$ So, yes, the final answer comes out to be $$P(\text{Pair}) = \frac{2}{5}\cdot\frac{1}{3}+\frac{2}{5}\cdot\frac{1}{3}+\frac{2}{5} \cdot\frac{1}{3} = \frac{2}{5}.$$ $\endgroup$
    – Em.
    Nov 29 '15 at 3:51

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