TIFR CSE 2010 | Part A | Question: 10

Question

A drawer contains $2$ Blue, $4$ Red and $2$ Yellow balls. No two balls have the same radius. If two balls are randomly selected from the drawer, what is the probability that they will be of the same colour?  

• A. $\left(\dfrac{2}{7}\right)$  
• B. $\left(\dfrac{2}{5}\right)$  
• C. $\left(\dfrac{3}{7}\right)$    
• D. $\left(\dfrac{1}{2}\right)$    
• E. $\left(\dfrac{3}{5}\right)$  

Answer 1

If any $2$ balls selected from $8$ balls then we can choose ${^8}C{_2}$ ways=$28$ ways

If selected $2$ balls are  same color then$ {^2}C{_2} + {^4}C{_2} +{^2}C{_2}$ ways=$1+6+1$ ways=$8$ ways

So, required probability=$\dfrac{8}{28}=\dfrac{2}{7}$

Correct Answer: $A$

Comments

I am wondering what would be the answer if the statement ' no two balls have the same radius ' is omitted..

Is it be (3/20)??

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The answer still remains same 2/7

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I agree to Rishabh gupta we need to arrange them also, it's just not  a selection.

Answer 2

Answer Should be.... 2/7

More initiative way to approach this.

Pr(Selected two balls will be of same color)= Pr(picking 2 blue balls or 2 red balls or 2 yellow balls)

Pr( 2Blue balls ) = 2/8 * 1/7

Pr( 2Red balls ) = 4/8 * 3/7

Pr( 2Yellow balls ) = 2/8 * 1/7

We get = 2/8 * 1/7 + 4/8 * 3/7 + 2/8 * 1/7 = 2/7