An urn contains nine balls, of which three are red, four are blue and

1.	An urn contains nine balls, of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is(a) \(\frac{1}{3}\)(b) \(\frac{2}{7}\)(c) \(\frac{1}{21}\)(d) \(\frac{2}{23}\)
Answer» (b) \(\frac{2}{7}\) Let S be the sample space having 9 balls (3R + 4B + 2G) Then n(S) = Total number of ways in which 3 balls can be drawn out of the 9 balls = \(\frac{9\times8\times7}{3\times2}\) = 84 Let A : Event of drawing three balls of different colours from the urn. ⇒ A = Event of drawing 1 red ball out of 3 red balls, 1 blue ball out of 4 blue balls and 1 green ball out of 2 green balls ⇒ n(A) = ³C₁ × ⁴C₁ × ²C₁ = 3 × 4 × 2 = 24. ∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{24}{84}\) = \(\frac{2}{7}\).

An urn contains nine balls, of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is(a) \(\frac{1}{3}\)(b) \(\frac{2}{7}\)(c) \(\frac{1}{21}\)(d) \(\frac{2}{23}\)

Answer»

(b) \(\frac{2}{7}\)

Let S be the sample space having 9 balls (3R + 4B + 2G)

Then n(S) = Total number of ways in which 3 balls can be drawn out of the 9 balls

= \(\frac{9\times8\times7}{3\times2}\) = 84

Let A : Event of drawing three balls of different colours from the urn.

⇒ A = Event of drawing 1 red ball out of 3 red balls, 1 blue ball out of 4 blue balls and 1 green ball out of 2 green balls

⇒ n(A) = ³C₁ × ⁴C₁ × ²C₁ = 3 × 4 × 2 = 24.

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{24}{84}\) = \(\frac{2}{7}\).

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