Question:
If \(x\) is chosen at random from the set \(\{1,2,3,4\}\) and \(y\) is chosen at random from the set \(\{5,6,7\}\), then the probability that \(xy\) will be even is
If \(x\) is chosen at random from the set \(\{1,2,3,4\}\) and \(y\) is chosen at random from the set \(\{5,6,7\}\), then the probability that \(xy\) will be even is
Show Hint
A product is odd only when all multiplying numbers are odd. Use the complement method to simplify such probability problems.
Updated On: Jun 15, 2026
- \(\dfrac{5}{6}\)
- \(\dfrac{1}{6}\)
- \(\dfrac{1}{2}\)
- \(\dfrac{2}{3}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Find total possible outcomes.
Set for \(x\) is
\[ \{1,2,3,4\} \] which contains \(4\) elements.
Set for \(y\) is
\[ \{5,6,7\} \] which contains \(3\) elements.
Therefore, total possible ordered pairs are
\[ 4\times3=12 \]
Step 2: Understand when \(xy\) is even.
The product \(xy\) is even if at least one of \(x\) or \(y\) is even.
Instead of counting directly, count the complement event.
\[ xy \text{ is odd} \] only when both \(x\) and \(y\) are odd.
Step 3: Count odd values.
Odd elements in \(\{1,2,3,4\}\):
\[ 1,3 \] So, number of odd choices for \(x\) is \(2\).
Odd elements in \(\{5,6,7\}\):
\[ 5,7 \] So, number of odd choices for \(y\) is \(2\).
Hence, number of outcomes where \(xy\) is odd:
\[ 2\times2=4 \]
Step 4: Find the probability.
Number of outcomes where \(xy\) is even:
\[ 12-4=8 \]
Therefore, required probability is
\[ \frac{8}{12} \]
\[ =\frac23 \]
Step 5: Final conclusion.
Hence, the probability that \(xy\) is even is
\[ \boxed{\frac23} \]
Set for \(x\) is
\[ \{1,2,3,4\} \] which contains \(4\) elements.
Set for \(y\) is
\[ \{5,6,7\} \] which contains \(3\) elements.
Therefore, total possible ordered pairs are
\[ 4\times3=12 \]
Step 2: Understand when \(xy\) is even.
The product \(xy\) is even if at least one of \(x\) or \(y\) is even.
Instead of counting directly, count the complement event.
\[ xy \text{ is odd} \] only when both \(x\) and \(y\) are odd.
Step 3: Count odd values.
Odd elements in \(\{1,2,3,4\}\):
\[ 1,3 \] So, number of odd choices for \(x\) is \(2\).
Odd elements in \(\{5,6,7\}\):
\[ 5,7 \] So, number of odd choices for \(y\) is \(2\).
Hence, number of outcomes where \(xy\) is odd:
\[ 2\times2=4 \]
Step 4: Find the probability.
Number of outcomes where \(xy\) is even:
\[ 12-4=8 \]
Therefore, required probability is
\[ \frac{8}{12} \]
\[ =\frac23 \]
Step 5: Final conclusion.
Hence, the probability that \(xy\) is even is
\[ \boxed{\frac23} \]
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