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?

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.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

the answer is 21%.