how can you find the probability of r tails in n tosses of a fair coin??

Answer 1

Dear me,

The binomial distribution gives you what you want.

Let f(r;n,p) be the probability of getting exactly r successes (i.e., tails) in exactly n trials (i.e., tosses), with probability p of success on any individual trial. This is written as:

f(r;n,p) = C(n,r) p^r (1 - p)^(n - r),

where C(n,r) = n! / [r! (n - r)!] is the binomial coefficient, often called "n choose r" because it gives the number of combinations for choosing exactly r items from a set of n items.

In your problem, p = 0.5, which is the usual definition of a fair coin. This gives you the following by substituting for p.

f(r;n,0.5) = C(n,r) 0.5^r (1 - 0.5)^(n - r)
= C(n,r) 0.5^r (0.5)^(n - r)
= C(n,r) 0.5^(r + n - r)
= C(n,r) 0.5^n .

For example, if you want the probability of exactly 2 tails in 3 tosses of a fair coin, you would have r = 2, n = 3, and p = 0.5, so

f(2;3,0.5) = C(3,2) 0.5^3
= 3! / [2! (3 - 2)!] (0.125)
= (3)(2)(1) / [(2)(1) (1)] (0.125)
= 6 / [2] (0.125)
= 3 (0.125)
= 0.375 (or 3/8).

Answer 2

Thats easy, its always .50
1/2 = tails/ heads
two sides to a coin and you have a 50% chance of getting tails

Answer 3

if the coin is fair then the chances of getting a tails would be 50% cause there is only two sides then you would multiply your n term by 50% which the equasion would be .5n=r cause 50% would be equal to .5

Answer 4

n*(1/2) = r

probability of r tails in n tosses = r/n = 1/2

Answer 5

the probabilty of r in n would be easier to solve for me if it had some numbers involved, i dont really know how to answer this without all the information.
if u say that i will be tossing a quarter wat is the probability that it will land on heads it would have a 1 to 2 ratio because both sides of the quarter are different.
srry if this doesnt help, or is confusing