Answer :
To solve this problem, we need to determine the probability of two independent events happening together: the coin showing tails and the cube showing either a three or a five.
First, let's find the probability of getting tails on the coin flip:
- A coin has two sides: heads and tails.
- The probability of flipping tails is [tex]\( \frac{1}{2} \)[/tex].
Next, let's find the probability of rolling a three or a five on a standard six-sided die:
- A standard die has six faces, numbered from 1 to 6.
- The favorable outcomes for rolling a three or a five are 3 and 5.
- There are two favorable outcomes out of six possible outcomes.
- Therefore, the probability of rolling a three or a five is [tex]\( \frac{2}{6} = \frac{1}{3} \)[/tex].
Now, to find the combined probability of both events happening (coin shows tails and die shows either a three or a five), we multiply the probabilities of the two independent events:
[tex]\[ P(\text{tails and (three or five)}) = P(\text{tails}) \times P(\text{three or five}) \][/tex]
[tex]\[ P(\text{tails and (three or five)}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \][/tex]
Hence, the probability that the coin will show tails and the cube will show a three or five is:
[tex]\[ \frac{1}{6} \][/tex]
First, let's find the probability of getting tails on the coin flip:
- A coin has two sides: heads and tails.
- The probability of flipping tails is [tex]\( \frac{1}{2} \)[/tex].
Next, let's find the probability of rolling a three or a five on a standard six-sided die:
- A standard die has six faces, numbered from 1 to 6.
- The favorable outcomes for rolling a three or a five are 3 and 5.
- There are two favorable outcomes out of six possible outcomes.
- Therefore, the probability of rolling a three or a five is [tex]\( \frac{2}{6} = \frac{1}{3} \)[/tex].
Now, to find the combined probability of both events happening (coin shows tails and die shows either a three or a five), we multiply the probabilities of the two independent events:
[tex]\[ P(\text{tails and (three or five)}) = P(\text{tails}) \times P(\text{three or five}) \][/tex]
[tex]\[ P(\text{tails and (three or five)}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \][/tex]
Hence, the probability that the coin will show tails and the cube will show a three or five is:
[tex]\[ \frac{1}{6} \][/tex]
