Chapter 10.8, Problem 15IP

(a)

To determine

The probability in which has the same number on both sides.

Answer to Problem 15IP

The probability in which has the same number on both sides is $\frac{6}{60}\text{or}\frac{1}{10}$ .

Explanation of Solution

Given information:

A game requires in tossing a 10 sided numbered solid and 6 sided numbered solid is moved on the game board.

Formula Used:

  $\text{P=}\frac{\text{No}\text{.}\text{of}\text{favorable}\text{outcomes}}{\text{No}\text{.}\text{of}\text{possible}\text{outcomes}}$

Calculation:

There are ten outcomes for the 10-sided die, and six outcomes for the 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 both be ones, both twos, both threes, both fours, both fives, and both sixes. So, the probability is $\frac{6}{60}\text{or}\frac{1}{10}$

Conclusion:

The probability in which has the same number on both sides is $\frac{6}{60}\text{or}\frac{1}{10}$ .

(b)

To determine

The probability of getting the value of (odd, even) or (even, odd)

Answer to Problem 15IP

The probability of getting the value of (odd, even) or (even, odd) is $\frac{30}{60}\text{or}\frac{1}{2}$

Explanation of Solution

Given information:

A game requires in tossing a 10 sided numbered solid and 6 sided numbered solid is moved on the game board.

Formula Used:

  $\text{P=}\frac{\text{No}\text{.}\text{of}\text{favorable}\text{outcomes}}{\text{No}\text{.}\text{of}\text{possible}\text{outcomes}}$

Calculation:

There are ten outcomes for the 10-sided die, and six outcomes for the 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 3 even numbers on the 6-sided die.

  $5\times 3=15$

There are 15 ways to get (odd, even). There are five even numbers on the 10-sided die, and 3 odd numbers on the 6-sided die.

  $5\times 3=15$

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

Therefore, there are 15 + 15 or 30 different ways to (odd, even) or (even, odd). So, the probability is $\frac{30}{60}\text{or}\frac{1}{2}$