Question

An unbiased die is tossed twice. What is the probability of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss?

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

Hint:

For independent events, multiply favorable outcomes of each eventDivide by total outcomes to get probability.

Answer 1

The Correct Option is D

Step 1: Total outcomes.
A die is thrown twice, so total outcomes are:
\[ 6 \times 6 = 36 \]

Step 2: Favorable outcomes for first toss.
First toss should be 4, 5 or 6.
Number of favorable outcomes:
\[ 3 \]

Step 3: Favorable outcomes for second toss.
Second toss should be 1, 2, 3 or 4.
Number of favorable outcomes:
\[ 4 \]

Step 4: Total favorable outcomes.
\[ 3 \times 4 = 12 \]

Step 5: Compute probability.
\[ P = \frac{12}{36} \]

Step 6: Simplify.
\[ P = \frac{1}{3} \]

Step 7: Final conclusion.
\[ \boxed{\frac{1}{3}} \]