Question

a coin is tossed and a die is rolled. find the probability of getting a tail and a number greater than 4. (type an integer or a simplified fraction.)

Explanation:

Step1: Calculate the probability of getting a tail

When a coin is tossed, there are 2 possible outcomes (head or tail). The probability of getting a tail, \(P(T)\), is \(\frac{1}{2}\) since \(n(T) = 1\) (number of tail - outcomes) and \(n(\text{total coin - outcomes})=2\). So, \(P(T)=\frac{1}{2}\).

Step2: Calculate the probability of getting a number greater than 4 on a die

When a die is rolled, there are 6 possible outcomes (\(1,2,3,4,5,6\)). The numbers greater than 4 are 5 and 6. So, \(n(\text{numbers}>4)=2\) and \(n(\text{total die - outcomes}) = 6\). The probability of getting a number greater than 4, \(P(N>4)\), is \(\frac{2}{6}=\frac{1}{3}\).

Step3: Use the multiplication rule for independent events

Since the coin - toss and die - roll are independent events, the probability of both events occurring is \(P(T\cap N>4)=P(T)\times P(N > 4)\). Substitute \(P(T)=\frac{1}{2}\) and \(P(N>4)=\frac{1}{3}\) into the formula: \(P(T\cap N>4)=\frac{1}{2}\times\frac{1}{3}\).

$$ \frac{1}{2}\times\frac{1}{3}=\frac{1\times1}{2\times3}=\frac{1}{6} $$

Answer:

\(\frac{1}{6}\)