Question

what is the probability of one of the coins landing on tails and two of them landing on heads?
1/4
3/8
1/2
3/4

Explanation:

Step1: Count total outcomes

From the tree - diagram, the total number of possible outcomes when flipping 3 coins is $n = 8$ (since there are 8 end - points in the tree - diagram: (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T), (H,H,H), (H,H,T)).

Step2: Count favorable outcomes

The favorable outcomes (one tail and two heads) are (H,T,H), (T,H,H), (H,H,T). So the number of favorable outcomes $m=3$.

Step3: Calculate probability

The probability $P$ of an event is given by the formula $P=\frac{m}{n}$. Substituting $m = 3$ and $n = 8$, we get $P=\frac{3}{8}$.

Answer:

$\frac{3}{8}$