Question

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

Answer 1

Answer on Question #54325-Math-Statistics and Probability

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

Solution

For 3 tosses, possibilities are:

$$0 \text{ heads} \rightarrow 3 \text{ tails} \rightarrow W = (\text{heads} - \text{tails}) = 0 - 3 = -3$$ $$1 \text{ head} \rightarrow 2 \text{ tails} \rightarrow W = (\text{heads} - \text{tails}) = 1 - 2 = -1$$ $$2 \text{ heads} \rightarrow 1 \text{ tail} \rightarrow W = (\text{heads} - \text{tails}) = 2 - 1 = 1$$ $$3 \text{ heads} \rightarrow 0 \text{ tails} \rightarrow W = (\text{heads} - \text{tails}) = 3 - 0 = 3$$

So, the elements of the sample space are:

$$S = \{-3, -1, 1, 3\}.$$

To find the probability for each case:

$$P(0 \text{ heads} \& 3 \text{ tails}) = \frac{3!}{0!(3 - 0)!} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^3 = \frac{1}{8}$$ $$P(1 \text{ head} \& 2 \text{ tails}) = \frac{3!}{1!(3 - 1)!} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = \frac{3}{8}$$ $$P(2 \text{ heads} \& 1 \text{ tail}) = \frac{3!}{2!(3 - 2)!} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^1 = \frac{3}{8}$$ $$P(3 \text{ heads} \& 0 \text{ tails}) = \frac{3!}{3!(3 - 3)!} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^0 = \frac{1}{8}$$

i.e.:

$$\text{For } W = -3 \rightarrow P = \frac{1}{8}$$ $$\text{For } W = -1 \rightarrow P = \frac{3}{8}$$ $$\text{For } W = 1 \rightarrow P = \frac{3}{8}$$ $$\text{For } W = 3 \rightarrow P = \frac{1}{8}$$

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