$Let:\ H_1$ be the event of at least one head

$Let:\ H_2$ be the event of at least two heads

$Let:\ T_1$ be the event of at least one Tail

a) at least one toss is fixed is a head

The problem is asking for $P(H_2|H_1)$; we will use the unconditioned probabilities to get the conditional probability

because the possible outcomes are {HHH, HHT, HTT, HTH, THH, THT, TTT, TTH}

$P(H_2) = 4/8$

$P(H_1) = 7/8$

$P(H_2|H_1) = P(H_1 \cap H_2) / P(H_1)$; by definition of conditional probability

• If you know that there are at least 2 heads, you automatically know that there is at least one head, so $H_2 \subset H_1$. Therefore, $H_1 \cap H_2 = H_2$.

$P(H_2|H_1) = P(H_2)/ P(H_1)$

$P(H_2|H_1) = 4/8 \cdot 8/7 = \frac{4}{7}$

We can verify this by looking at the outcomes in H_1 and counting the ones with at least two heads: $\{HHH, HHT, HTT, HTH, THH, THT, TTH\} \rightarrow \frac{4}{7}$

b) If at least one toss is a TAIL

This problem is asking for $P(H_2|T_1)$; Use the logic above to get $P(T_1)$

$P(T_1) = 7/8$

$P(H_2|T_1) = P(T_1 \cap H_2) / P(T_1)$; by conditional probability

• We cannot use independence to get the numerator on the right side, because we do not know that $T_1$ and $H_2$ are independent. Instead, we count the outcomes in $T_1 \cap H_2$:

$$\{THH, HTH, HHT\}$$

There are three outcomes, because we know that there are two heads and one tail in three positions, and there are three places that the tail can go. Therefore,

$$P(T_1 \cap H_2) = 3/8$$

Plugging in above,

$$P(H_2|T_1) = \frac{3/8}{7/8} = \frac{3}{7}$$

We can verify this by looking at the event $T_1$ and counting how may outcomes include two heads: $\{TTT, HHT, HTT, HTH, THH, THT, TTH\} \rightarrow \frac{3}{7}$