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(N).

Short answer

The value of$P\left(N\right)is0.36$.

Step-by-step solution

Given Information

In a package, we have 12 hats, 15 noisemakers, 10 finger traps, and 5 confetti bags.

Assume.

H stands for the occasion of receiving a hat.

N denotes the probability of receiving a noisemaker.

F = the occurrence of a finger trap.

C = the event of receiving a confetti bag.

Concept Used

Probability is a metric that indicates how confident we are in the results of a certain experiment. The probability is calculated using the following formula:

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

If we flip a coin twice, for example, the sample space associated with this random experiment is $\{HH,HT,TH,TT\}$where T= tails and H= heads. Assume that A= receives one tail. There are two possible outcomes that are favourable to the event. A

localid="1648969007925" $\{HT,TH\}\text{, so}P\left(A\right)=\frac{2}{4}$

$=0.5\text{.}$

Calculation

To calculate the likelihood of receiving a noisemaker, multiply the favourable number of cases by 15 and the overall number of cases by 42.

As a result, your chances of receiving a noisemaker are:

$P\left(N\right)=\frac{15}{42}$

$=\frac{5}{14}$

$=0.36$