If 4 marbles are drawn at random, what is the probability that 2 are red and 2 are blue ?

Solution

Number of ways of drawing 4 marbles out of 16

=> $$n(S) = C^{16}_4 = \frac{16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4}$$

= $$1820$$

Out of the four drawn marbles, 2 are red and 2 are blue.

=> $$n(E) = C^8_2 \times C^3_2 = \frac{8 \times 7}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}$$

= $$28 \times 3 = 84$$

$$\therefore$$ Required probability = $$\frac{n(E)}{n(S)}$$

= $$\frac{84}{1820} = \frac{3}{65}$$

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