A bag contains 4 red and 6 black balls. Three balls are drawn at random. If x denotes the number of red balls, then
Match List-I with List-II.
| List-I | List-II | ||
| (Probability) | (Value) | ||
| (A) | P (x=0) | (I) | \( \displaystyle\frac{1}{30} \) |
| (B) | P (x=1) | (II) | \( \displaystyle\frac{3}{10} \) |
| (C) | P (x=2) | (III) | \( \displaystyle\frac{1}{2} \) |
| (D) | P (x=3) | (IV) | \( \displaystyle\frac{1}{6} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
Three balls can be chosen from 10 in \( {}^{10}C_{3}=120 \) ways. For \( r \) red balls the number of favourable ways is \( {}^{4}C_{r}\times{}^{6}C_{3-r} \).
(A) P (x=0)
This matches (IV) \( \frac{1}{6} \).
\( \displaystyle P(x=0)=\frac{{}^{6}C_{3}}{120}=\frac{20}{120}=\frac{1}{6} \)
(B) P (x=1)
This matches (III) \( \frac{1}{2} \).
\( \displaystyle P(x=1)=\frac{{}^{4}C_{1}\times{}^{6}C_{2}}{120}=\frac{4\times 15}{120}=\frac{1}{2} \)
(C) P (x=2)
This matches (II) \( \frac{3}{10} \).
\( \displaystyle P(x=2)=\frac{{}^{4}C_{2}\times{}^{6}C_{1}}{120}=\frac{6\times 6}{120}=\frac{3}{10} \)
(D) P (x=3)
This matches (I) \( \frac{1}{30} \).
\( \displaystyle P(x=3)=\frac{{}^{4}C_{3}}{120}=\frac{4}{120}=\frac{1}{30} \)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).