Out of a collection of hats which contains red and blue hat 2 hats are drawn at random. The number of hats belong to the set {1,2,3,4............13).The probability that they have different color is $\frac{1}{2}$ .Possible number of blue hats is?

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Solution

The correct option is A
3

Let number of red and blue hat be X and Y.

$P (E) = \frac{X \times Y}{[^{(X + Y)} C_{2}]} = \frac{1}{2}$

$= (X + Y) (X + Y - 1) = 4 X Y$

$X^{2} + Y^{2} + 2 X Y - X - Y = 4 X Y$

$X^{2} + Y^{2} + 2 X Y - X - Y = 0$

$Y^{2} - Y (2 X + 1) + (X^{2} - X) = 0$

$D = (2 X + 1)^{2} - 4 (X^{2} - X) = 8 X + 1$ has to be perfect square

hence possible values are 1,3,6,10

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Out of a collection of hats which contains red and blue hat 2 hats are drawn at random. The number of hats belong to the set {1,2,3,4............13).The probability that they have different color is 12.Possible number of blue hats is?

The correct option is A 3 Let number of red and blue hat be X and Y. P(E)=X× Y[(X+Y)C2]=12 =(X+Y)(X+Y−1)=4XY X2+Y2+2XY−X−Y=4XY X2+Y2+2XY−X−Y=0 Y2−Y(2X+1)+(X2−X)=0 D=(2X+1)2−4(X2−X)=8X+1 has to be perfect square hence possible values are 1,3,6,10

Out of a collection of hats which contains red and blue hat 2 hats are drawn at random. The number of hats belong to the set {1,2,3,4............13).The probability that they have different color is $\frac{1}{2}$ .Possible number of blue hats is?