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GATE 2025 CH – Question 14

Engineering Mathematics · Probability and Statistics: Probability, conditional probability, descriptive statistics · 1 mark · Multiple choice

A box contains 3 identical green balls and 7 identical blue balls. Two balls are randomly drawn without replacement from the box. The probability of drawing 1 green and 1 blue ball is

  1. $\frac{{}^3P_1 \times {}^7P_1}{{}^{10}P_2}$
  2. $\frac{{}^{10}P_3 \times {}^{10}P_7}{{}^{10}P_2}$
  3. $\frac{{}^{10}C_3 \times {}^{10}C_7}{{}^{10}C_2}$
  4. $\frac{{}^3C_1 \times {}^7C_1}{{}^{10}C_2}$

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Correct answer: (D) $\frac{{}^3C_1 \times {}^7C_1}{{}^{10}C_2}$

Explanation

The number of ways to pick any 2 balls out of 10 is $^{10}C_2$. The number of ways to pick 1 of the 3 green balls and 1 of the 7 blue balls is $^3C_1 \times {}^7C_1$. The probability is $\frac{{}^3C_1 \times {}^7C_1}{{}^{10}C_2} = \frac{21}{45}$. The options with permutations count ordered selections only in the numerator or the denominator, and so are not the correct ratio.