Question

A die is rolled twice independently. Probability that either first die shows number no less than \(4\) or second die shows at least \(4\)?

• A. \(\frac{1}{2}\)
• B. \(\frac{5}{6}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{3}{4}\)

Hint:

For independent events, \[ P(A\cap B)=P(A)P(B). \] For an ``either-or'' probability, always use \[ P(A\cup B)=P(A)+P(B)-P(A\cap B). \]

Answer 1

The Correct Option is D

Concept: When a probability question contains the word ``or'', we use \[ P(A\cup B)=P(A)+P(B)-P(A\cap B). \] This avoids double counting of outcomes that satisfy both events simultaneously.

Step 1: Define the events.
Let \[ A=\{\text{first die shows }4,5,6\} \] and \[ B=\{\text{second die shows }4,5,6\}. \] Therefore, \[ P(A)=\frac{3}{6}=\frac12, \qquad P(B)=\frac{3}{6}=\frac12. \]

Step 2: Find the probability of intersection.
Since the two throws are independent, \[ P(A\cap B) = P(A)P(B) = \frac12\cdot\frac12 = \frac14. \]

Step 3: Apply the addition theorem.
\[ P(A\cup B) = \frac12+\frac12-\frac14 = \frac34. \] Hence, \[ \boxed{\frac34} \]