Question

two six - sided dice are tossed.
event a: the first die does not land on 1.
event b: the second die does not land 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

Explanation:

Step1: Find P(A)

A six - sided die has 6 possible outcomes. Event A is the first die not landing on 1. So the number of favorable outcomes for A is \(6 - 1=5\). The probability \(P(A)=\frac{\text{Number of favorable outcomes for A}}{\text{Total number of outcomes}}=\frac{5}{6}\).

Step2: Find P(B)

Event B is the second die not landing on 6. The number of favorable outcomes for B is \(6 - 1 = 5\). The probability \(P(B)=\frac{\text{Number of favorable outcomes for B}}{\text{Total number of outcomes}}=\frac{5}{6}\).

Step3: Find P(A and B)

Since the two dice tosses are independent events, we use the formula \(P(A\text{ and }B)=P(A)\times P(B)\). Substitute \(P(A)=\frac{5}{6}\) and \(P(B)=\frac{5}{6}\) into the formula: \(P(A\text{ and }B)=\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}\).
To convert this to a percentage, we calculate \(\frac{25}{36}\approx0.6944\), and then multiply by 100: \(0.6944\times100 = 69.44\%\approx69\%\) (rounded to the nearest whole number).

Answer:

69