You toss a coin twice. Which calculation proves that landing on tails for the first toss and heads on the second toss are independent events?

Answer with explanation:

If a coin is tossed twice than the sample space of outcomes are:

{ HH, HT, TH, TT}

Now let A denote the event that the tail comes up in the first toss

and B denote the event that head comes up on the second toss.

Let P denote the probability of an event.

Hence,

P (A) = 1/2 (As probability of tail is half during single toss)

Similarly P (B) = 1/2 (Since, Probability of obtaining a head is half on a single toss)

Also,

P (A∩B) = 1/4

(Since the outcomes is TH)

Now, as:

P (A∩B) = P (A) * P (B)

Hence, landing on tails for the first toss and heads on the second toss are independent events.

i dont completely understand the question so if im wrong im sorry but i think the answer is a 1out 2 percent chance if it lands on tails some with heads