Two cubes have their face painted as either red or blue color. 1st cube has 5 faces red and 1 face blue. The Probability that both the faces show same color on rolling is $\frac{1}{2}$ . How many red faces are present on the $2^{n d}$ cube.

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Solution

The correct option is C 3
Suppose x faces of cube 2 is red then 6-x will be blue.

P(RR or BB)= $\frac{5}{6} \times \frac{x}{6} + \frac{1}{6} \times \frac{(6 - x)}{6} = \frac{1}{2}$

$\frac{5 x}{36} + \frac{(6 - x)}{36} = \frac{1}{2}$

4x+6=18

x=3

Hence number of red faces is 3.

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Two cubes have their face painted as either red or blue color. 1st cube has 5 faces red and 1 face blue. The Probability that both the faces show same color on rolling is 12. How many red faces are present on the 2nd cube.

The correct option is C 3Suppose x faces of cube 2 is red then 6-x will be blue. P(RR or BB)= 56×x6+16×(6−x)6=12 5x36+(6−x)36=12 4x+6=18 x=3 Hence number of red faces is 3.

Two cubes have their face painted as either red or blue color. 1st cube has 5 faces red and 1 face blue. The Probability that both the faces show same color on rolling is $\frac{1}{2}$ . How many red faces are present on the $2^{n d}$ cube.

The correct option is C 3
Suppose x faces of cube 2 is red then 6-x will be blue.

P(RR or BB)= $\frac{5}{6} \times \frac{x}{6} + \frac{1}{6} \times \frac{(6 - x)}{6} = \frac{1}{2}$

$\frac{5 x}{36} + \frac{(6 - x)}{36} = \frac{1}{2}$

4x+6=18

x=3

Hence number of red faces is 3.