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 \(P(H)\).

Short answer

\(P(H)=0.29\)

Step-by-step solution

Step 1. 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.

Step 2. 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 hat, the favorable number of cases is \(12\) and total cases are \(42\). Therefore, the probability of getting a hat is:

\(P(H)=\frac{12}{42}=\frac{2}{7}=0.29\)

\(P(H)=0.29\)