Answer:
31/32
Explanation:
If a coin is tossed 5 times, obtaining at least 1 tail means you can either obtain 1,2,3,4, or 5 tails.
[tex]$\text{P(getting at least one tail in 5 tosses)}$=1-P(5\text{ heads in 5 tosses)}[/tex]In a coin toss:
[tex]\begin{gathered} P(\text{obtaining head)}=\frac{1}{2} \\ P(\text{obtaining tails)}=\frac{1}{2} \end{gathered}[/tex]Therefore:
[tex]P(5\text{ heads in 5 tosses)}=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=(\frac{1}{2})^5=\frac{1}{32}[/tex]This then means that the probability of getting at least 1 tail:
[tex]$\text{P(getting at least one tail in 5 tosses)}$=1-\frac{1}{32}=\frac{31}{32}[/tex]The probability is 31/32.
Note: You can also solve this problem using the binomial probability formula.