Question

A coin is tossed 8 times. What is the probability of getting all heads? Express your answer as a simplified fraction or a decimal rounded to four decimal places.

Answer 1

Answer:

The probability of getting all heads is $\frac{1}{256}$.

Step-by-step explanation:

For each time the coin is tossed, there are only two possible outcomes. Either it is heads, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

A coin is tossed 8 times. This means that $n = 8$

In each coin toss, heads or tails are equally as likely. So $p = \frac{1}{2}$

What is the probability of getting all heads?

This is $P(X = 8)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$[tex]P(X = 8) = C_{8,8}*(\frac{1}{2})^{8}*(1 - \frac{1}{2})^{0}$ = $\frac{1}{256}$

The probability of getting all heads is $\frac{1}{256}$.