Question

Given a gear bag with 3 bungee cords, 5 socket wrenches, and 2 hammers, what is the probability a person will randomly draw a socket wrench from the bag? Group of answer choice

Answer 1

Answer:

$\frac{1}{2}$

Step-by-step explanation:

Given:

A gear bag with 3 bungee cords, 5 socket wrenches, and 2 hammers.

Total number of instruments = 3 + 5 + 2 = 10

Question asked:

What is the probability a person will randomly draw a socket wrench from the bag?

Solution:

As we know:

$Probability =\frac{Favourable \ outcome}{Total\ outcome}$

First of all we will find favorable outcome for drawing 1 socket wrench from the bag randomly:

Favorable outcome for drawing 1 socket wrench out of  cookies 3 bungee cords =$^{3} C_{0}$

Favorable outcome for drawing 1 socket wrench out of 5 socket wrenches = $^{5} C_{1}$

Favorable outcome for drawing 1 socket wrench out of 2 hammers = $^{2} C_{0}$

Thus, total Favorable outcome for drawing 1 socket wrench from the bag randomly: By using:

$^{n} C_{r}=\frac{n!}{(n-r)!\ r!}$

$^{3} C_{0}$ $\times$$^{5} C_{1}$$\times$ $^{2} C_{0}$

$\frac{3!}{(3-0)!\ 0!} \times\frac{5!}{(5-1)!\ 1!}\times\frac{2!}{(2-0)!\ 0!}$

$\frac{3!}{3!\times1} \times\frac{5!}{4!\times1} \times\frac{2!}{2!\times1} \\\\ 1\times\frac{5\times4!}{4!} \times\frac{2!}{2!} \\\\ 1\times5\times1=5$

Total outcome for drawing 1 socket wrench out of 10 pieces of instruments:

$^{10} C_{1} =\frac{10!}{(10-1)!\times1!} =\frac{10\times9!}{9!} =10$

Now,

$Probability =\frac{Favourable \ outcome}{Total\ outcome}$

                  $=\frac{5}{10} =\frac{1}{2}$

Thus, the probability that a person will randomly draw a socket wrench from the bag is $\frac{1}{2}$