A bag contains 10 balls out of which k are red and (10?k) are black, where . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black ba

Question

A bag contains 10 balls out of which k are red and (10−k) are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is: [2026]

Smart Achievers · Accepted Answer

(4)

$Probability = \frac{{}^{1}C_{0} \cdot {}^{9}C_{3}}{\sum_{k = 0}^{10} {}^{k}C_{0} \cdot {}^{10 - k}C_{3}}$

$= \frac{{}^{9}C_{3}}{{}^{10}C_{3} + {}^{9}C_{3} + {}^{8}C_{3} + \dots + {}^{3}C_{3}}$

$= \frac{{}^{9}C_{3}}{{}^{11}C_{4}}$

$= \frac{14}{55}$

Solution

Q.
A bag contains 10 balls out of which k are red and (10−k) are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is: [2026]

1 $\frac{7}{110}$

2 $\frac{7}{55}$

3 $\frac{7}{11}$

4 $\frac{14}{55}$

Ans.
(4)

$Probability = \frac{{}^{1}C_{0} \cdot {}^{9}C_{3}}{\sum_{k = 0}^{10} {}^{k}C_{0} \cdot {}^{10 - k}C_{3}}$

$= \frac{{}^{9}C_{3}}{{}^{10}C_{3} + {}^{9}C_{3} + {}^{8}C_{3} + \dots + {}^{3}C_{3}}$

$= \frac{{}^{9}C_{3}}{{}^{11}C_{4}}$

$= \frac{14}{55}$

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Solution

Q. A bag contains 10 balls out of which k are red and (10−k) are black, where 0≤k≤10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is: [2026]

1 7110

2 755

3 711

4 1455

Ans. (4) Probability=C01·C39∑k=010C0k·C310-k =C39C310+C39+C38+⋯+C33 =C39C411 =1455