Question

An unfair coin has probability 0.3 of landing heads. The coin is tossed seven times. What is the probability that it lands heads at least once? Round the answer to four decimal places.

Answer 1

Answer:

0.9176 = 91.76% probability that it lands heads at least once

Step-by-step explanation:

For each time that the coin is tossed, there are only two possible outcomes. Either it lands heads, or it does not. The probability of a toss landing heads is independent of other tosses. So we use the binomial probability distribution to solve this question.

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.

An unfair coin has probability 0.3 of landing heads.

This means that $p = 0.3$

The coin is tossed seven times.

This means that $n = 7$

What is the probability that it lands heads at least once?

Either it does not lands heads, or it lands at least once. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \geq 1)$. So

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

In which

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

$P(X = 0) = C_{7,0}.(0.3)^{0}.(0.7)^{7} = 0.0824$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0824 = 0.9176$

0.9176 = 91.76% probability that it lands heads at least once