Math / Data & Statistics

QuestionA box contains 15 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below. a. All are defective b. None are defective a. The probability is $\square$ (Type a fraction. Simplify your answer.)

Studdy Solution

STEP 1

What is this asking? We're picking 4 transistors out of a box of 15, and 4 of those 15 are broken.
We want to know the chances of picking all broken ones, and the chances of picking all working ones. Watch out! Don't mix up the number of broken transistors with the number we're picking out!
Also, remember that probabilities are always between 0 and 1.

STEP 2

1. Calculate the total number of ways to choose 4 transistors.
2. Calculate the number of ways to choose 4 defective transistors.
3. Calculate the probability of choosing 4 defective transistors.
4. Calculate the number of ways to choose 4 working transistors.
5. Calculate the probability of choosing 4 working transistors.

STEP 3

We're choosing 4 transistors out of a total of 15.
The order doesn't matter, so we use combinations.
The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items and $k$ is the number we're choosing.

STEP 4

In our case, $n = 15$ and $k = 4$ , so we have $\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15!}{4!11!} = \frac{15 \cdot 14 \cdot 13 \cdot 12}{4 \cdot 3 \cdot 2 \cdot 1} = 15 \cdot 7 \cdot 13 \cdot \frac{12}{12 \cdot 2} = 1365$ .
There are 1365 ways to choose any 4 transistors out of 15.

STEP 5

There are 4 defective transistors, and we want to choose all 4 of them.
The number of ways to do this is $\binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = \frac{4!}{4! \cdot 1} = 1$ .
Makes sense, right?
There's only one way to pick all the broken ones!

STEP 6

The probability is the number of ways to choose 4 defective transistors divided by the total number of ways to choose any 4 transistors.
So, the probability is $\frac{1}{1365}$ .

STEP 7

If 4 out of the 15 transistors are defective, then $15 - 4 = 11$ are working.
We want to choose 4 working transistors out of the 11 available.
This is $\binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 11 \cdot 10 \cdot 3 \cdot \frac{8}{8} = 330$ .
There are 330 ways to pick 4 working transistors.

STEP 8

The probability is the number of ways to choose 4 working transistors divided by the total number of ways to choose any 4 transistors.
So, the probability is $\frac{330}{1365} = \frac{330}{1365} = \frac{22}{91}$ .

STEP 9

a. The probability of selecting 4 defective transistors is $\frac{1}{1365}$ . b. The probability of selecting 4 working transistors is $\frac{22}{91}$ .

Was this helpful?

QuestionA box contains 15 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below. a. All are defective b. None are defective a. The probability is $\square$ (Type a fraction. Simplify your answer.)

Studdy Solution

STEP 1

STEP 2

STEP 3

We're choosing 4 transistors out of a total of 15.
The order doesn't matter, so we use combinations.
The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items and $k$ is the number we're choosing.

STEP 4

In our case, $n = 15$ and $k = 4$ , so we have $\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15!}{4!11!} = \frac{15 \cdot 14 \cdot 13 \cdot 12}{4 \cdot 3 \cdot 2 \cdot 1} = 15 \cdot 7 \cdot 13 \cdot \frac{12}{12 \cdot 2} = 1365$ .
There are 1365 ways to choose any 4 transistors out of 15.

STEP 5

There are 4 defective transistors, and we want to choose all 4 of them.
The number of ways to do this is $\binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = \frac{4!}{4! \cdot 1} = 1$ .
Makes sense, right?
There's only one way to pick all the broken ones!

STEP 6

The probability is the number of ways to choose 4 defective transistors divided by the total number of ways to choose any 4 transistors.
So, the probability is $\frac{1}{1365}$ .

STEP 7

STEP 8

The probability is the number of ways to choose 4 working transistors divided by the total number of ways to choose any 4 transistors.
So, the probability is $\frac{330}{1365} = \frac{330}{1365} = \frac{22}{91}$ .

STEP 9

a. The probability of selecting 4 defective transistors is $\frac{1}{1365}$ . b. The probability of selecting 4 working transistors is $\frac{22}{91}$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionA box contains 15 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below. a. All are defective b. None are defective a. The probability is □\square□ (Type a fraction. Simplify your answer.)

Studdy Solution

STEP 1

STEP 2

STEP 3

We're choosing 4 transistors out of a total of **15**.The order doesn't matter, so we use combinations.The formula for combinations is (nk)=n!k!(n−k)! \binom{n}{k} = \frac{n!}{k!(n-k)!} (kn​)=k!(n−k)!n!​, where nnn is the **total number of items** and kkk is the **number we're choosing**.

STEP 4

STEP 5

STEP 6

The probability is the number of ways to choose 4 defective transistors divided by the total number of ways to choose any 4 transistors.So, the probability is 11365\frac{1}{1365}13651​.

STEP 7

STEP 8

The probability is the number of ways to choose 4 working transistors divided by the total number of ways to choose any 4 transistors.So, the probability is 3301365=3301365=2291\frac{330}{1365} = \frac{330}{1365} = \frac{22}{91}1365330​=1365330​=9122​.

STEP 9

a. The probability of selecting 4 defective transistors is 11365\frac{1}{1365}13651​. b. The probability of selecting 4 working transistors is 2291\frac{22}{91}9122​.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionA box contains 15 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below. a. All are defective b. None are defective a. The probability is $\square$ (Type a fraction. Simplify your answer.)

We're choosing 4 transistors out of a total of 15.
The order doesn't matter, so we use combinations.
The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items and $k$ is the number we're choosing.

The probability is the number of ways to choose 4 defective transistors divided by the total number of ways to choose any 4 transistors.
So, the probability is $\frac{1}{1365}$ .

The probability is the number of ways to choose 4 working transistors divided by the total number of ways to choose any 4 transistors.
So, the probability is $\frac{330}{1365} = \frac{330}{1365} = \frac{22}{91}$ .

a. The probability of selecting 4 defective transistors is $\frac{1}{1365}$ . b. The probability of selecting 4 working transistors is $\frac{22}{91}$ .