Chapter 13.8, Problem 13PPS

i.

To determine

Probabilities of the given conditions.

Answer to Problem 13PPS

  $\frac{6}{60}$ or $\frac{1}{10}$ is probability required for the given condition.

Explanation of Solution

Given:

  $10-$ side and $6-$ sided die is tossed to determine how to move on a game board.

Condition:

  $P$ (same number on both dice)

Concept Used:

Fundamental counting principle.

Calculation:

There are ten outcomes for $10-$ sided die and six outcomes for $6-$ sided die.

  $10\times 6=60$

So, the number of possible outcomes is $60$ .

There are six ways that the toss results in the same number on both dice. They could be ones, both twos, both threes, both fours, both fives and both sixes.

So the probability is $\frac{6}{60}$ or $\frac{1}{10}$ .

Conclusion:

Thus, Probabilities of the given conditions is $\frac{6}{60}$ or $\frac{1}{10}$ .

ii.

To determine

Probabilities of the given conditions.

Answer to Problem 13PPS

  $\frac{30}{60}$ or $\frac{1}{2}$ is probability required for the given condition.

Explanation of Solution

Given:

  $10-$ side and $6-$ sided die is tossed to determine how to move on a game board.

Condition:

  $P$ (odd, even) or $P$ (even, odd)

Concept Used:

Fundamental counting principle.

Calculation:

There are ten outcomes for $10-$ sided die and six outcomes for $6-$ sided die.

  $10\times 6=60$

So, the number of possible outcomes is $60$ .

There are five odd numbers on the $10-$ sided die and three even numbers on the $6-$ sided die.

  $\begin{array}{l}=5\times 3\\ =15\end{array}$

There are $15$ ways to get (odd, even).

There are five even numbers on the $10-$ sided die and three odd numbers on the $6-$ sided die.

  $\begin{array}{l}=5\times 3\\ =15\end{array}$

There are $15$ ways to get (even, odd).

Thus, there are total $15+15=30$ different ways to (odd, even) or (even, odd).

Probability is $\frac{30}{60}$ or $\frac{1}{2}$ .

Conclusion:

Thus, Probabilities of the given conditions is $\frac{30}{60}$ or $\frac{1}{2}$ .