Chapter 10.8, Problem 17IP

(a)

To determine

The probability for getting atleast one 2.

Answer to Problem 17IP

The probability of getting atleast one 2 is $\frac{7}{16}$ .

Explanation of Solution

Given information:

The spinners are at the right is spun once. The tree diagram is used to determine the probabilities as,

  

Formula Used:

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

Calculation:

A tree diagram would give the following possible outcomes:

  $2-1,2-2,2-3,2-4,5-1,5-2,5-3,5-4,6-1,6-2,6-3,\text{ and }6-4$

In order to keep the probabilities the same, $6-1,6-2,6-3,\text{ and }6-4$ must be counted twice.

So, there are 16 possible outcomes.

There are 7 outcomes where there is at least one 2. So, P (at least one 2) is $\frac{7}{16}$ .

Conclusion:

The probability of getting atleast one 2 is $\frac{7}{16}$

(b)

To determine

The probability for getting atleast one 3.

Answer to Problem 17IP

The probability for getting atleast one 3 is $\frac{4}{16}\text{or}\frac{1}{4}$

Explanation of Solution

Given information:

The spinners are at the right is spun once. The tree diagram is used to determine the probabilities as,

  

Formula Used:

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

Calculation:

A tree diagram would give the following possible outcomes:

  $2-1,2-2,2-3,2-4,5-1,5-2,5-3,5-4,6-1,6-2,6-3,\text{ and }6-4$

In order to keep the probabilities the same, $6-1,6-2,6-3,\text{ and }6-4$ must be counted twice.

So, there are 16 possible outcomes.

There are 4 outcomes where there is at least one 3.

So, P (at least one 3) is $\frac{4}{16}\text{or}\frac{1}{4}$

Conclusion:

The probability for getting atleast one 3 is $\frac{4}{16}\text{or}\frac{1}{4}$