Question

A box is filled with several party favors. It contains 12

hats, 15 noisemakers, ten finger traps, and five bags of confetti.

Let H = the event of getting a hat.

Let N = the event of getting a noisemaker.

Let F = the event of getting a finger trap.

Let C = the event of getting a bag of confetti.

Find P(F).

Short answer

$P\left(F\right)=\frac{10}{42}=\frac{5}{21}=0.24$

Step-by-step solution

Given Information

In the given question, we are given the following information:

A box contains 12 hats, 15 noisemakers, 10 finger traps, and 5 bags of confetti.

Concept Used

Probability is a measure that is associated with how certain we are of outcomes of a particular experiment.

The formula for calculating the probability is:

Probability $=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

For example, if we flip a coin two times, the sample space associated with this random experiment is

$\{HH,HT,TH,TT\}$where T= tails and H= heads . Let's suppose A= getting one tail. There are two

outcomes which favors the event A

$\{HT,TH\}\text{, so}P\left(A\right)=\frac{2}{4}=0.5\text{.}$

Calculation

Let H= the event of getting a hat.

Let N= the event of getting a noisemaker.

Let F= the event of getting a finger trap.

Let C= the event of getting a bag of confetti.

Now to find the probability of getting a finger trap, the favorable number of cases is 10 and total cases are 42 .

Therefore, the probability of getting a finger trap is:

$P\left(F\right)=\frac{10}{42}=\frac{5}{21}=0.24$

Conclusion

$P\left(F\right)=0.24$