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Probability question

2010 · Shift 0 · Q31
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Probability question

2010 · Shift 0 · Q31

JEE MainMathematicsProbabilityMCQ+4 / −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 :
  1. A
    27{2 \over 7}72​
  2. B
    121{1 \over 21}211​
  3. C
    123{1 \over 23}231​
  4. D
    13{1 \over 3}31​
View written solutionFree

Correct answer: A

  1. Total number of ways to draw 3 balls from 9

Since 3 balls are drawn without replacement and order does not matter,

Total ways=(93)=84.\text{Total ways} = {9 \choose 3} = 84.Total ways=(39​)=84.
  1. Favourable cases: all three balls have different colours

The urn has:

  • 3 red balls
  • 4 blue balls
  • 2 green balls

For the three balls to have different colours, we must choose:

  • 1 red from 3,
  • 1 blue from 4,
  • 1 green from 2.

So,

Favourable ways=(31)(41)(21)=3⋅4⋅2=24.\text{Favourable ways} = {3 \choose 1}{4 \choose 1}{2 \choose 1} = 3 \cdot 4 \cdot 2 = 24.Favourable ways=(13​)(14​)(12​)=3⋅4⋅2=24.
  1. Required probability
P(different colours)=2484=27.P(\text{different colours}) = \frac{24}{84} = \frac{2}{7}.P(different colours)=8424​=72​.
  1. Option check
  • A: 27\dfrac{2}{7}72​ ✅
  • B: 121\dfrac{1}{21}211​ ❌
  • C: 123\dfrac{1}{23}231​ ❌
  • D: 13\dfrac{1}{3}31​ ❌

Therefore, the correct answer is:

27\boxed{\frac{2}{7}}72​​
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