Three balls are drawn one by one without replacement from a bag containing 5 white and 4 red balls. Let X denote the number of white balls.
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | P(X = 0) | I. | 10/21 |
| B. | P(X = 1) | II. | 5/42 |
| C. | P(X = 2) | III. | 1/21 |
| D. | P(X = 3) | IV. | 5/14 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-I, D-II
Drawing 3 balls one by one without replacement is the same as choosing 3 of the 9 balls together.
\( n(S)={}^{9}C_{3}=84 \)
\( \displaystyle P(X=r)=\frac{{}^{5}C_{r}\cdot{}^{4}C_{3-r}}{84} \)
(A) P(X = 0)
This matches III. 1/21.
\( \displaystyle P(X=0)=\frac{{}^{4}C_{3}}{84}=\frac{4}{84}=\frac{1}{21} \)
(B) P(X = 1)
This matches IV. 5/14.
\( \displaystyle P(X=1)=\frac{5\times{}^{4}C_{2}}{84}=\frac{30}{84}=\frac{5}{14} \)
(C) P(X = 2)
This matches I. 10/21.
\( \displaystyle P(X=2)=\frac{{}^{5}C_{2}\times 4}{84}=\frac{40}{84}=\frac{10}{21} \)
(D) P(X = 3)
This matches II. 5/42.
\( \displaystyle P(X=3)=\frac{{}^{5}C_{3}}{84}=\frac{10}{84}=\frac{5}{42} \)
So the matching is A-III, B-IV, C-I, D-II.
NCERT Exemplar:
Probability
