A bag contains 5 green and 7 red balls, out of which two balls are dra

1.	A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ?(a) \(\frac{5}{11}\)(b) \(\frac{7}{22}\)(c) \(\frac{16}{33}\)(d) \(\frac{31}{66}\)
Answer» (d) \(\frac{31}{66}\) Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls. Then n(S) = ¹²C₂ = \(\frac{12\times11}{2}\) = 66 ∴ Let A : Event of drawing two red balls ⇒ n(A) = ⁷C₂ = \(\frac{7\times6}{2}\) = 21 B : Event of drawing two green balls ⇒ n(B) = ⁵C₂ = \(\frac{5\times4}{2}\) = 10 ∴ P(Event of drawing 2 balls of same colour) = P(Drawing two red balls) or P(drawing two green balls) = P(A) + P(B) = \(\frac{n(A)}{n(S)}\) + \(\frac{n(B)}{n(S)}\) = \(\frac{21}{66}\) + \(\frac{10}{66}\) = \(\frac{31}{66}\).

A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ?(a) \(\frac{5}{11}\)(b) \(\frac{7}{22}\)(c) \(\frac{16}{33}\)(d) \(\frac{31}{66}\)

Answer»

(d) \(\frac{31}{66}\)

Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.

Then

n(S) = ¹²C₂ = \(\frac{12\times11}{2}\) = 66

∴ Let A : Event of drawing two red balls

⇒ n(A) = ⁷C₂ = \(\frac{7\times6}{2}\) = 21

B : Event of drawing two green balls

⇒ n(B) = ⁵C₂ = \(\frac{5\times4}{2}\) = 10

∴ P(Event of drawing 2 balls of same colour) = P(Drawing two red balls) or P(drawing two green balls)

= P(A) + P(B) = \(\frac{n(A)}{n(S)}\) + \(\frac{n(B)}{n(S)}\) = \(\frac{21}{66}\) + \(\frac{10}{66}\) = \(\frac{31}{66}\).

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