Question

let t be a random variable giving the number of heads plus the number of tails in three tosses of a coin.list the elements of the sample space s for the three tosses of the coin and assign a value to each sample point.

Answer 1

Answer on Question #71243 – Math – Statistics and Probability

Question

Let $t$ be a random variable giving the number of heads plus the number of tails in three tosses of a coin. List the elements of the sample space $S$ for the three tosses of the coin and assign a value to each sample point.

Solution

The sample space $S$ for the three tosses of the coin is:

$$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$

For 3 tosses, possibilities are:

$$HHH: 3 \text{ heads and } 0 \text{ tails} \Rightarrow t = (\text{heads} + \text{tails} = 3 + 0 = 3)$$ $$HHT: 2 \text{ heads and } 1 \text{ tail} \Rightarrow t = (\text{heads} + \text{tails} = 2 + 1 = 3)$$ $$HTH: 2 \text{ heads and } 1 \text{ tail} \Rightarrow t = (\text{heads} + \text{tails} = 2 + 1 = 3)$$ $$HTT: 1 \text{ head and } 2 \text{ tails} \Rightarrow t = (\text{heads} + \text{tails} = 1 + 2 = 3)$$ $$THH: 2 \text{ heads and } 1 \text{ tail} \Rightarrow t = (\text{heads} + \text{tails} = 2 + 1 = 3)$$ $$THT: 1 \text{ head and } 2 \text{ tails} \Rightarrow t = (\text{heads} + \text{tails} = 1 + 2 = 3)$$ $$TTH: 1 \text{ head and } 2 \text{ tails} \Rightarrow t = (\text{heads} + \text{tails} = 1 + 2 = 3)$$ $$TTT: 0 \text{ heads and } 3 \text{ tails} \Rightarrow t = (\text{heads} + \text{tails} = 0 + 3 = 3)$$$$\begin{array}{l} P(t = 3) = P(HHH) + P(HHT) + P(HTH) + P(HTT) + P(THH) + P(THT) + \\ + P(TTH) + P(TTT) = \\ = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \\ + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = 8 \left(\frac{1}{8}\right) = 1 \end{array}$$

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