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The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

The probability of getting a head or a tail is $\frac{1}{2}$and both are independent events. So, the probability of getting first 3 heads followed by 3 tails is obtained as follows,

$$\begin{gathered} p={\left(\frac{1}{2}\right)}^3脳{\left(\frac{1}{2}\right)}^3 \\ ={\left(\frac{1}{2}\right)}^6 \end{gathered}$$

Find the probability of hh.

$\frac{2}{3}脳\frac{2}{3}=\frac{4}{9}$

Find the probability of ht.

$\frac{2}{3}脳\frac{1}{3}=\frac{2}{9}$

Find the probability of th.

$\frac{1}{3}脳\frac{2}{3}=\frac{2}{9}$
Find the probability of .

$\frac{1}{3}脳\frac{1}{3}=\frac{1}{9}$

Set the sample space and their associated probability.

It can be observed that the sum of all the probabilities is 1.

From the obtained sample space and the probabilities, the desired probability, that is $P\left(A\right)=\frac{8}{9}$.

Find the probability of two heads if you know there was at least one head.

$$\begin{gathered} P_A\left(B\right)=\frac{{\displaystyle \frac{4}{9}}}{{\displaystyle \frac{8}{9}}} \\ =\frac{1}{2} \end{gathered}$$
Thus, the probability of at least one head is $\frac{8}{9}$and the probability of two heads if you know there was at least one head, $P_A\left(B\right)=\frac{1}{2}$.

The possibility is 8 in number, namely hhh, hht, hth, htt, thh, tht, tth, ttt .

Set the sample space and their associated probability.

When there are more heads, this implies that the sample space becomes hhh,hht,hth,thh and each point has a probability of $\frac{1}{4}$and the probability of having one tail is $\frac{3}{4}$.

When no two head appear in succession, this implies that the sample space becomes hth,htt,tht,tth,ttt and each point has a probability of $\frac{1}{5}$and the probability of having all tail is $\frac{1}{5}$.

When coins do not fall all alike, this implies that the sample space becomes hht,hth,htt,thh,tht,tth and each point has a probability of $\frac{1}{6}$. The event that two coins fell in succession are hht,htt,thh,tth and the probability that two coins fall were in succession is $\frac{4}{6}$or $\frac{2}{3}$.

The sample space is hhh,hht,hth,htt,thh,tht,tth,ttt and contains 8 points and each point has a probability of $\frac{1}{8}$.

The difference of number of heads and tails is 1 when either number of head is 2 or number of tails is 2. The event for the condition is hht, hth, htt, thh, tht, tth.

The probability that the difference of number of heads and tails is 1 is $\frac{4}{6}$or $\frac{2}{3}$.

When there is at least one head, this implies that the sample space becomes hhh, hht, hth, htt, thh, tht, tth and each point has a probability of $\frac{1}{7}$.

The probability of having exactly 2 heads is $\frac{3}{7}$.

Thus, the probability of one tail when there were more heads than tails is $\frac{3}{4}$, the probability of all tails when two heads did not appear in succession is $\frac{1}{5}$, the probability that two coins were in succession were alike, when the coins did not all fall alike is $\frac{2}{3}$, the probability that the difference of number of heads and tails is 1 is $\frac{2}{3}$, the probability of exactly two heads when there was at least one head is $\frac{3}{7}$.