Question

A boy has a collection of blue and green marbles. The number of blue marbles belong to the sets $2,3,4,13$. If two marbles are chosen simultaneously and at random from his collection, then the probability that they have different colour is $1/2$. Possible number of blue marbles is:

• A. $2$
• B. $3$
• C. $6$
• D. $10$

Answer 1

The correct option is B $3$

Suppose, there are m blue and n green marbles.

There are (m+n)*(m+n−1)/2(m+n2)=(m+n)(m+n−1)2 ways to choose 22 marbles.

There are m*nmn ways to choose 22 marbles with different colours.

The probability of getting two marbles with different colours is therefore

2mn(m+n)(m+n−1)2mn(m+n)(m+n−1)

So, the probability is 1212, if and only if

(m+n)(m+n−1)=4mn(m+n)(m+n−1)=4mn

1) m=2 : n2+3n+2=8n1) m=2 : n2+3n+2=8n has no solution in NN

2) m=3 : n2+5n+6=12n2) m=3 : n2+5n+6=12n has the solutions 11 and 66.

3) m=6 : n2+11n+30=24n3) m=6 : n2+11n+30=24n has the solutions 33 and 1010.

4) m=10 : n2+19n+90=40n4) m=10 : n2+19n+90=40n has the solutions 66 and 1515.