Two socks are to be picked at random from a drawer containing only bla

Question

Two socks are to be picked at random from a drawer containing only black and white socks. What is the probability that both are white?

(1) The probability of the first sock being black is 1/3.
(2) There are 24 white socks in the drawer.

Skywalker18 · Answer

We need to find Probability that both are white

(1) The probability of the first sock being black is 1/3.
Probability (First sock is black) = 1/3
Probability (First sock is white) = 1 - 1/3 = 2/3
But unless we know the total number of socks in bag , we can't determine the probability that both are white . 
Not sufficient

(2) There are 24 white socks in the drawer. 
Number of white socks = 24
Not sufficient

Combining 1 and 2 , we get
Probability (First sock is white) = 1 - 1/3= 2/3
Number of white socks = 24
Total number of socks = 3/2 * 24
= 36
Now we can calculate the probability that both are white .
Answer C

LogicGuru1 · Answer

(1) The probability of the first sock being black is  \frac{1}{3} .
It means the probabilty of first white sock is  \frac{2}{3} 
This is a unique statements because it gives us the ratio of black to white socks as 1:2

Number of Black socks = x

Number of White socks = 2x

Number of Total socks = 3x

Probability of White =  \frac{2x}{3x} = \frac{2}{3} 
 
after taking one White sock the number of Black socks remain unchanged= x

after taking one White socks the number of White socks = 2x -1

after taking one white sock the number of Total socks = 3x -1

after taking one White sock the number of Probability of Picking another white =  \frac{2x-1}{3x-1}

Total probablity of picking two white =  \frac{2x}{3x} *\frac{2x-1}{3x-1}

Since we dont know total number of socks X INSUFFICIENT

(2) There are 24 white socks in the drawer
We do not know how many black socks are there in the drawer so we cannot use this information 
INSUFFICIENT

MERGE BOTH 
since ratio of black to white is 1:2 therefore we know that there are total 12 black socks in the drawer 
Now we can easily find the probablity of picking two white socks

P(W,W)=  \frac{24}{36}*\frac{23}{35}

ANSWER IS C

fskilnik · Answer

From the question stem and statement (1) we adopt the following:

(*) ASSUMPTION: We are dealing with two sequential extractions without replacement.

? = P\left( {2\,\,white\,\,from\,\,\left( * ight)} ight)

\left( 1 ight)\,\,{	ext{black}}\,\,:\,\,{	ext{white}}\,\,\, = \,\,\,1:2\,\,\,\,\,\,\left\{ \begin{gathered}
 \,Take\,\,\left( {b,w} ight) = \left( {1,2} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}\,\,\,\, \hfill \
 \,Take\,\,\left( {b,w} ight) = \left( {2,4} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = \frac{4}{6} \cdot \frac{3}{5} 
e \frac{1}{3} \hfill \ 
\end{gathered} ight.

\left. {\left( 2 ight)\,\,{	ext{white}} = 24\,\,\,\,\left\{ \begin{gathered}
 \,Take\,\,\left( {b,w} ight) = \left( {1,24} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,?\,\, \cong \,\,\,1\,\,\,\, \hfill \
 \,Take\,\,\left( {b,w} ight) = \left( {{{10}^6},24} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? \cong 0 \hfill \ 
\end{gathered} ight.\,\,\,} ight\}\,\,\,\,\,{	ext{extremal}}\,\,{	ext{evaluations}}\,

\left( {1 + 2} ight)\,\,\,\,\left\{ \begin{gathered}
 black = 12 \hfill \
 {	ext{white}} = 24 \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = \,\,{	ext{unique}}\,\,\,\,\,\,\,\,\,\left( {\frac{{24}}{{36}} \cdot \frac{{23}}{{35}}} ight)

The above follows the notations and rationale taught in the GMATH method.

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Two socks are to be picked at random from a drawer containing only bla

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