From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let X denotes the number of defective items in the sample. If the sample is drawn randomly, then Match List-I with List-II (P(X) denotes the probability of occurrence of X)
| List-I | List-II |
| (A) P (X = 0) | (I) \( \displaystyle\frac{^{3}C_{3}\times{}^{7}C_{1}}{^{10}C_{4}} \) |
| (B) P (X = 1) | (II) \( \displaystyle\frac{^{7}C_{4}\times{}^{3}C_{0}}{^{10}C_{4}} \) |
| (C) P (X = 2) | (III) \( \displaystyle\frac{^{3}C_{1}\times{}^{7}C_{3}}{^{10}C_{4}} \) |
| (D) P (X = 3) | (IV) \( \displaystyle\frac{^{3}C_{2}\times{}^{7}C_{2}}{^{10}C_{4}} \) |
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
X defectives in a sample of 4 means choosing X of the 3 defective items and the other 4 − X from the 7 good ones:
\( \displaystyle P(X=r)=\frac{^{3}C_{r}\times{}^{7}C_{4-r}}{^{10}C_{4}} \)
(A) P (X = 0)
This matches (II). Take r = 0: \( {}^{3}C_{0}\times{}^{7}C_{4} \) in the numerator.
(B) P (X = 1)
This matches (III). Take r = 1: \( {}^{3}C_{1}\times{}^{7}C_{3} \).
(C) P (X = 2)
This matches (IV). Take r = 2: \( {}^{3}C_{2}\times{}^{7}C_{2} \).
(D) P (X = 3)
This matches (I). Take r = 3: \( {}^{3}C_{3}\times{}^{7}C_{1} \).
So the matching is (A) - (II), (B) - (III), (C) - (IV), (D) - (I).