Question

a coin is tossed three times. an outcome is represented by a string of the sort htt (meaning a head on the first toss, followed by two tails). the 8 outcomes are listed in the table below. note that each outcome has the same probability. for each of the three events in the table, check the outcome(s) that are contained in the event. then, in the last column, enter the probability of the event. event a: alternating head and tail (with either coming first) event b: two or more heads event c: exactly one head outcomes hhh tht hth tth htt hht thh ttt probability

Explanation:

Step1: Identify total number of outcomes

When a coin is tossed 3 times, the total number of possible outcomes is $2\times2\times2 = 8$ since each toss has 2 possibilities (head or tail).

Step2: Analyze Event A

Event A: Alternating head and tail (with either coming first). The outcomes for Event A are THT and HTH. So $n(A)=2$. The probability of Event A, $P(A)=\frac{n(A)}{n(S)}=\frac{2}{8}=\frac{1}{4}$.

Step3: Analyze Event B

Event B: Two or more heads. The outcomes for Event B are HHH, HHT, HTH, THH. So $n(B) = 4$. The probability of Event B, $P(B)=\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}$.

Step4: Analyze Event C

Event C: Exactly one head. The outcomes for Event C are HTT, THT, TTH. So $n(C)=3$. The probability of Event C, $P(C)=\frac{n(C)}{n(S)}=\frac{3}{8}$.

Outcomes	Event A	Event B	Event C
THT	$\checkmark$		$\checkmark$
HTH	$\checkmark$	$\checkmark$
TTH			$\checkmark$
HTT			$\checkmark$
HHT		$\checkmark$
THH		$\checkmark$
TTT
Probability	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{3}{8}$

Answer:

Outcomes	Event A	Event B	Event C
THT	$\checkmark$		$\checkmark$
HTH	$\checkmark$	$\checkmark$
TTH			$\checkmark$
HTT			$\checkmark$
HHT		$\checkmark$
THH		$\checkmark$
TTT
Probability	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{3}{8}$