Question

two six - sided dice are tossed.
event a: the first die lands on 1, 2, 3, or 4.
event b: the second die lands on 6.
what is the probability that both events will occur?
p(a and b) =? %
for independent events: p(a and b) = p(a)·p(b)
round your answer to the nearest wha

Explanation:

Step1: Calculate P(A)

A die has 6 sides. Event A is the first die landing on 1, 2, 3, or 4. So the number of favorable outcomes for A is 4. The probability \( P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total number of outcomes}} = \frac{4}{6}=\frac{2}{3}\).

Step2: Calculate P(B)

Event B is the second die landing on 6. The number of favorable outcomes for B is 1. The probability \( P(B) = \frac{\text{Number of favorable outcomes for B}}{\text{Total number of outcomes}}=\frac{1}{6}\).

Step3: Calculate P(A and B)

Since the two dice tosses are independent events, we use the formula \( P(A \text{ and } B)=P(A)\times P(B) \). Substituting the values of \( P(A) \) and \( P(B) \) we found: \( P(A \text{ and } B)=\frac{2}{3}\times\frac{1}{6}=\frac{2}{18}=\frac{1}{9}\approx 0.1111 \). To convert this to a percentage, we multiply by 100: \( 0.1111\times100 = 11.11\% \), and rounding to the nearest whole number gives \( 11\% \).

Answer:

11%