Question:
A bag contains \(7\) green and \(5\) black balls. \(3\) balls are drawn at random one after the other. If the balls are not replaced, then the probability of all three balls being green is
A bag contains \(7\) green and \(5\) black balls. \(3\) balls are drawn at random one after the other. If the balls are not replaced, then the probability of all three balls being green is
Show Hint
In probability problems without replacement, multiply the successive probabilities because each draw changes the total number of objects.
Updated On: Jun 15, 2026
- \(\dfrac{343}{1720}\)
- \(\dfrac{21}{36}\)
- \(\dfrac{12}{35}\)
- \(\dfrac{7}{44}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Find the total number of balls.
The bag contains
\[ 7 \text{ green balls} \] and
\[ 5 \text{ black balls} \]
Therefore, total balls are
\[ 7+5=12 \]
Step 2: Find the probability of drawing green balls successively.
Since the balls are drawn without replacement, the probabilities change after each draw.
Probability that the first ball is green:
\[ \frac{7}{12} \]
After drawing one green ball, remaining green balls \(=6\) and total balls \(=11\).
Probability that the second ball is green:
\[ \frac{6}{11} \]
Now remaining green balls \(=5\) and total balls \(=10\).
Probability that the third ball is green:
\[ \frac{5}{10} \]
Step 3: Multiply the probabilities.
Required probability is
\[ \frac{7}{12}\times\frac{6}{11}\times\frac{5}{10} \]
\[ =\frac{7\times6\times5}{12\times11\times10} \]
Simplifying,
\[ =\frac{7}{44} \]
Step 4: Final conclusion.
Hence, the probability that all three balls are green is
\[ \boxed{\frac{7}{44}} \]
The bag contains
\[ 7 \text{ green balls} \] and
\[ 5 \text{ black balls} \]
Therefore, total balls are
\[ 7+5=12 \]
Step 2: Find the probability of drawing green balls successively.
Since the balls are drawn without replacement, the probabilities change after each draw.
Probability that the first ball is green:
\[ \frac{7}{12} \]
After drawing one green ball, remaining green balls \(=6\) and total balls \(=11\).
Probability that the second ball is green:
\[ \frac{6}{11} \]
Now remaining green balls \(=5\) and total balls \(=10\).
Probability that the third ball is green:
\[ \frac{5}{10} \]
Step 3: Multiply the probabilities.
Required probability is
\[ \frac{7}{12}\times\frac{6}{11}\times\frac{5}{10} \]
\[ =\frac{7\times6\times5}{12\times11\times10} \]
Simplifying,
\[ =\frac{7}{44} \]
Step 4: Final conclusion.
Hence, the probability that all three balls are green is
\[ \boxed{\frac{7}{44}} \]
Was this answer helpful?
0
0
Top AP EAMCET Mathematics Questions
- If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
- If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
- If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
- If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
- If the line joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is a focal chord, then one possible values of $\cot \frac{\alpha}{2} . \cot \frac{\beta}{2}$ is
View More Questions
Top AP EAMCET Probability Questions
- Three numbers are chosen at random from 1 to 20. The probability that their sum is divisible by 3 is:
- Two persons A and B throw three unbiased dice one after the other. If A gets the sum 13, then the probability that B gets a higher sum is:
- 8 teachers and 4 students are sitting around a circular table at random. The probability that no two students sit together is:
- When an unfair dice is thrown, the probability of getting a number \( k \) on it is \( P(X = k) = k^2 P \), where \( k = 1, 2, 3, 4, 5, 6 \) and \( X \) is the random variable denoting a number on the dice. Then, the mean of \( X \) is:
- A box contains \(100\) balls, numbered from \(1\) to \(100\). If \(3\) balls are selected one after the other at random with replacement from the box, then the probability that the sum of the three numbers on the balls selected from the box is an odd number, is
View More Questions
Top AP EAMCET Questions
- If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
- If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
- If only $\frac{1}{51}^{th}$ ot the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1}{11}^{th}$ of the main voltage is to be developed across the galvanometer, then the resistance required is $R_2$. Then $\frac{R_2}{R_1} =$
- If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
- If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
View More Questions