Question:

Let two fair six-faced dice \(A\) and \(B\) be thrown simultaneously. If \(E_1\) is the event that die \(A\) shows up four, \(E_2\) is the event that die \(B\) shows up two and \(E_3\) is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

Show Hint

Check pairwise independence by comparing joint probabilities with products, then check the triple intersection.
Updated On: Nov 11, 2026
  • \(E_1\) and \(E_2\) are independent.
  • \(E_2\) and \(E_3\) are independent.
  • \(E_1\) and \(E_3\) are independent.
  • \(E_1,\ E_2\) and \(E_3\) are independent.
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

\(P(E_1)=\frac16,\ P(E_2)=\frac16,\ P(E_3)=\frac12\) (the sum is odd for 18 of the 36 outcomes).
\(E_1\cap E_2\): only (4,2), \(P=\frac1{36}=\frac16\cdot\frac16\). Independent.
\(E_2\cap E_3\): B shows 2 and A must be odd: 3 outcomes, \(P=\frac{3}{36}=\frac1{12}=\frac16\cdot\frac12\). Independent.
\(E_1\cap E_3\): A shows 4 and B must be odd: 3 outcomes, \(P=\frac1{12}=\frac16\cdot\frac12\). Independent.
\(E_1\cap E_2\cap E_3\): (4,2) has an even sum, so the intersection is empty and \(P=0\ne\frac16\cdot\frac16\cdot\frac12\). The three events are not mutually independent.
So the statement that is NOT true is (4).
Answer: (4) \(E_1,\ E_2\) and \(E_3\) are independent.
Was this answer helpful?
0
0