Question

Find the probability of each event.

A fair coin is flipped ten times. What is
the probability of the coin landing tails up
at least nine times?

Answer 1

Using the binomial distribution, it is found that there is a 0.0108 = 1.08% probability of the coin landing tails up at least nine times.

What is the binomial distribution formula?

The formula is:

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• The coin is fair, hence p = 0.5.

• The coin is tossed 10 times, hence n = 10.

The probability that is lands tails up at least nine times is given by:

$P(X \geq 9) = P(X = 9) + P(X = 10)$

In which:

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

$P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098$

$P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.001$

Hence:

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0098 + 0.001 = 0.0108$

0.0108 = 1.08% probability of the coin landing tails up at least nine times.

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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