Question

You toss a fair coin 5 times. What is the probability of at least one head? Round to the nearest ten- thousandth.

Answer 1

Answer:

0.9688

Step-by-step explanation:

For each time the coin is tossed, there are only two possible outcomes. Either it is heads, or it is tails. The probabilities for each coin toss are independent from each other. 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:

For each time the coin is tossed, heads or tails are equally as likely, since the coin is fair. So $p = \frac{1}{2} = 0.5$

You toss a fair coin 5 times. What is the probability of at least one head?

Either there are no heads, or there is at least one head. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want to find $P(X \geq 1)$, when $n = 5$.

So

$P(X \geq 1) = 1 - P(X = 0)$

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

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.03125 = 0.96875$

Rounding to the nearest ten-thousandth(four decimal places), this probability is 0.9688.