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

Step 1: Reframe with the complement.
Instead of using the addition rule, it is often cleaner to find the chance the event fails and subtract from $1$. The event is: first die at least $4$ OR second die at least $4$.
Step 2: Describe the failure case.
The event fails only when BOTH dice show less than $4$, that is both show one of $\{1,2,3\}$.
Step 3: Probability one die is below 4.
Each die shows $1,2,$ or $3$ with probability $\frac{3}{6}=\frac{1}{2}$.
Step 4: Combine the two dice.
Since the throws are independent, both below $4$ has probability $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.
Step 5: Subtract from one.
The wanted probability is $1-\frac{1}{4}=\frac{3}{4}$.
Step 6: Cross-check by counting.
Out of $36$ equally likely outcomes, $9$ have both dice in $\{1,2,3\}$, so $36-9=27$ favourable, and $\frac{27}{36}=\frac{3}{4}$. This matches option 4.
\[ \boxed{\frac{3}{4}} \]