Chapter 17, Problem 1CTB

To determine

To find: The most and the least probable outcome.

Answer to Problem 1CTB

Solution: All the three outcomes are equally likely, so from the provided outcomes the most and the least probable outcomes cannot be judged only by 10 trials.

Explanation of Solution

Calculation:

According to the question,

$\begin{array}{c}\text{Favorable outcome}=\left(\begin{array}{l}\text{Number of possibilities of}\\ \text{occuring of a single sequence}\end{array}\right)\\ =1\\ \text{Total outcomes}=\text{Total number of possible sequence}\\ ={\left(2\right)}^{10}\end{array}$

Compute the probability of getting any of the three sequences of 10 particular outcomes as follows:

$\begin{array}{c}P\left(\text{Getting any specific sequence}\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{1}{{\left(2\right)}^{10}}\\ =\frac{1}{1024}\end{array}$

Out of 10 different tosses of an unbiased coin, there is equal number of chances of getting any combinations of both head and tails, randomly, or getting only heads or tails, which is $\frac{1}{1024}$ as obtained above. In a sequence of only 10 trials, it is not possible to determine the most and the least probable outcome. If the number of trials would have been large then the outcomes may have followed a definite pattern and in such a case the most and the least probable outcome could be easily determined. Therefore, it can be concluded that the three outcomes are equally likely.