Question

A fair coin is flipped 3 times. Let X be the number of heads in the 3 tosses.

Answer 1

Answer:

1)

P(X = 0) = 0.125

P(X = 1) = 0.375

P(X = 2) = 0.375

P(X = 3) = 0.125

2)

E(X) = 1.5

Step-by-step explanation:

For each coin, there are only two possible outcomes. Either it is heads, or it is tails. The probability of a toss being heads or tails is independent of any other toss. This means that the binomial probability distribution is used 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.

The expected value of the binomial distribution is $E(X) = np$.

A fair coin is flipped 3 times.

Fair means that it is equally as likely to be heads or tails, so $p = 0.5$.

3 times means that $n = 3$

(1) Find the pmf for X;

Probability of each outcome. So

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

$P(X = 0) = C_{3,0}.(0.5)^{0}.(0.5)^{3} = 0.125$

$P(X = 1) = C_{3,1}.(0.5)^{1}.(0.5)^{2} = 0.375$

$P(X = 2) = C_{3,2}.(0.5)^{2}.(0.5)^{1} = 0.375$

$P(X = 3) = C_{3,3}.(0.5)^{3}.(0.5)^{0} = 0.125$

(2) calculate the expected value of X.

$E(X) = np = 3*0.5 = 1.5$