Probability MCQ 1: (Drawer, Different color socks)

Probability MCQ 1: In a drawer there are 4 white socks, 3 blue socks and 5 grey socks. Two socks are picked randomly. What is the possibility that both the socks are of same color?

a) 4/11                b) 1                        c) 2/33                        d) 19/66


Total socks of all colors = 4+3+5=12

When we take out two socks, we want these to be either both white, blue or grey!

We know that probability formula for an event is:

\[ P(event)=\frac{Specific Sample}{Total Sample} \]


Note: We read the notation “P(event)” as ‘probability of event’

Here when we draw first time, there will be a total sample of 12 socks out of which there are 4 white socks !

So probability for first time draw to be a white sock event is:


\[ P(1stWhiteSock)=\frac{4}{12}=\frac{1}{3} \]

For second time draw there will be left 11 socks as total sample and there will be 3 white socks!


So, probability for second draw to be a white sock event is:

\[ P(2ndWhiteSock)=\frac{3}{11}\]

So, total probability for white sock event is:

Since above each of two draws are separately performed or say, are independent of each other, we have total probability formula as under:


\[ P(1stWhiteSockAnd2ndWhiteSock)=P(1stWhiteSock)\times P(2ndWhiteSock) \]


\[ =\frac{1}{3}\times \frac {3}{11}=\frac{1}{11} \]

Similarly, for 3 blue socks in the total sample, we want both times draw blue socks!


\[ P(1stBlueSock)=\frac{3}{12}=\frac{1}{4} \]


\[ P(2nd Blue Sock)=\frac{2}{11}\]

Total probability for both blue socks is

\[ P(1stBlueSockAnd2ndBlueSock)=\frac{1}{4}\times \frac{2}{11}=\frac{1}{22} \]

Finally, for grey socks we have,

\[ P(1stGreySockAnd2ndGreySock)=\frac{5}{12}\times \frac{4}{11}=\frac{5}{33} \]


Since events of A= both socks being white,B=both socks being blue or C=Both sock being grey are mutually exclusive , we have probability formula for mutually exclusive events A,B and C as;

\[ P(A orBorC)=P(A)+P(B)+P(C) \]


Probability for both time the socks being white, blue or grey is;

\[ P(AorBorC)=\frac{1}{11}+\frac{1}{22}+\frac{5}{33}=\frac{19}{66} \]


So, the answer is ‘d’



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