A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn

1.	A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black. ?
Answer» Let A : Event of getting at least 3 black balls Then n(A) = ⁵C₃ x ¹¹C₁ + ⁵C₄ (∵ Besides 5 black balls, there are 11 other balls) (3 black + others) (4 black) = \(\frac{5\times4}{2}\) x 11 + 5 = 115 Total numbers of ways in which 4 balls can be drawn from (7 + 5 + 4) = 16 balls n(S) = ¹⁶C₄ = \(\frac{16\times15\times14\times13}{4\times3\times2\times1}\) = 1820 ∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{115}{1820}\) = \(\frac{23}{364}.\)

A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black. ?

Answer»

Let A : Event of getting at least 3 black balls

Then n(A) = ⁵C₃ x ¹¹C₁ + ⁵C₄

(∵ Besides 5 black balls, there are 11 other balls)

(3 black + others) (4 black)

= \(\frac{5\times4}{2}\) x 11 + 5 = 115

Total numbers of ways in which 4 balls can be drawn from (7 + 5 + 4) = 16 balls

n(S) = ¹⁶C₄ = \(\frac{16\times15\times14\times13}{4\times3\times2\times1}\) = 1820

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{115}{1820}\) = \(\frac{23}{364}.\)

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A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black. ?

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