Question

Let W be a random variable giving the number of heads minus the number of tails in FOUR tosses of a coin. List the elements of the sample space S for the FOUR tosses of the coin and to each sample point assign a value w of W.
b) Find the probability distribution in tabular form.
c) Find the probability mass function for W.

Answer 1

Answer on Question #83130 – Math – Statistics and Probability

Question

Let W be a random variable giving the number of heads minus the number of tails in FOUR tosses of a coin. List the elements of the sample space S for the FOUR tosses of the coin and to each sample point assign a value w of W.

b) Find the probability distribution in tabular form.

c) Find the probability mass function for W.

Solution

The elements of the sample space S for the FOUR tosses of the coin are:

$$W = \{-4, -2, 0, 2, 4\}$$

b) The probability distribution in tabular form:

Where $\mathbf{w}$ is a value of W, and $P(w)$ is probability of w

c) The probability mass function for W:

$$P(w) = \binom{n}{k} p^k (1 - p)^{n - k}$$

Where $\pmb{p}$ is probability of heads in ONE toss of a coin, $\pmb{p} = 1/2$, $(1-p) = 1/2$

$$\binom{n}{k} = \frac{n!}{k!(n-k)!},$$ $$k = \frac{w}{2} + 2, \quad n = 4.$$

Thus, $P(w) = \binom{n}{k} p^k (1 - p)^{n - k} = \frac{4!}{\binom{w}{2} + 2! \left(4 - \frac{w}{2} - 2\right)!} * \left(\frac{1}{2}\right)^4 = \frac{24}{16 * \left(2 + \frac{w}{2}\right)! \left(2 - \frac{w}{2}\right)!}$

Answer:

The elements of the sample space S for the FOUR tosses of the coin are:

$$W = \{-4, -2, 0, 2, 4\}$$

b) The probability distribution in tabular form:

c) The probability mass function for W:

$$P(w) = \frac{24}{16 \times \left(2 + \frac{w}{2}\right)! \left(2 - \frac{w}{2}\right)!}.$$

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