Question

Three coins are tossed once. Find the probability of getting all tails.

Answer 1

Step 1: Find the number of elements in the sample space

When three coins are tossed, the possible outcomes are $\left\{\mathrm{HHH},\mathrm{TTT},\mathrm{HTT},\mathrm{THT},\mathrm{TTH},\mathrm{THH},\mathrm{HTH},\mathrm{HHT}\right\}$.
The sample space is S $=\left\{\mathrm{HHH},\mathrm{TTT},\mathrm{HTT},\mathrm{THT},\mathrm{TTH},\mathrm{THH},\mathrm{HTH},\mathrm{HHT}\right\}$

Number of elements in sample space$,n\left(S\right)=8$

Step 2: Find the probability of getting all tails

$\because$Probability of getting an event $=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}$

The favorable outcome is $\left\{\mathrm{TTT}\right\}$

So, event space $\mathrm{E}=\left\{\mathrm{TTT}\right\}$

Let $\mathrm{E}$ denotes event of getting all tails.

$\therefore \mathrm{n}\left(\mathrm{E}\right)=1$

$\mathrm{P}(\mathrm{getting}\mathrm{all}\mathrm{tails})=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}$

$=\frac{1}{8}$

Therefore, the required probability is $\frac{1}{8}$