91Ó°ÊÓ

The given statement is:

A balanced dime is tossed three times. The possible outcomes can be represented as follows:

HHH, HTH, THH, TTH, HHT, HTT, THT, TTT

The total number of all outcomes with the last two tosses come up tails are:

HTT, TTT

They are 2 in total.

Now Let's take the occurrence 'E' as the last two tosses come up tails.

The probability that the last two tosses come up tails is:

$$\begin{gathered} P\left(E\right)=\frac{2}{8} \\ =0.25 \end{gathered}$$

The total number of all outcomes with all three tosses comes up the same. are:

TTT, HHH

They are 2 in total.

Now Let's take the occurrence 'E' as all three tosses come up the same.

The probability that all three tosses come up the same is:

$$\begin{gathered} P\left(E\right)=\frac{2}{8} \\ =0.25 \end{gathered}$$

The total number of all outcomes with the second toss comes up heads. are:

HHH, THH, HHT, THT

They are 4 in total.

Now Let's take the occurrence 'E' as the second toss comes up heads.

The probability that the second toss comes up heads is:

$$\begin{gathered} P\left(E\right)=\frac{4}{8} \\ =0.500 \end{gathered}$$