Math  /  Data & Statistics

Question12 (4 pts.) Of 50 iPads available for checkout at the library, 10 are defective. A sample of 3 iPads are randomly selected without replacement. What is the probability that at least one of the iPads is defective? Round to the nearest thousandth. Show your work.

Studdy Solution

STEP 1

What is this asking? Out of 50 iPads, 10 are broken.
If we grab 3 iPads, what are the chances at least one is broken? Watch out! It's easier to calculate the chance that *none* are broken, then subtract that from 1!

STEP 2

1. Calculate the probability of picking a working iPad.
2. Calculate the probability of picking three working iPads in a row.
3. Calculate the probability of picking at least one broken iPad.

STEP 3

We know there are a total of 5050 iPads.
If 1010 are defective, that means 5010=4050 - 10 = 40 are *working*.

STEP 4

The probability of grabbing a working iPad on our first try is the number of working iPads divided by the total number of iPads: 4050\frac{40}{50}.
Let's simplify this fraction by dividing both the top and bottom by 1010, which gives us 45\frac{4}{5}.
So there's a 45\frac{4}{5} chance of grabbing a working iPad first!

STEP 5

For the *second* iPad, there's now one less working iPad and one less iPad overall.
So the probability of grabbing *another* working iPad is 401501=3949\frac{40-1}{50-1} = \frac{39}{49}.

STEP 6

For the *third* iPad, we do the same thing!
The probability is 391491=3848\frac{39-1}{49-1} = \frac{38}{48}.
Let's simplify this fraction by dividing the top and bottom by 22, giving us 1924\frac{19}{24}.

STEP 7

To get the probability of *all three* events happening, we multiply the individual probabilities together: 4539491924=4391954924=29645880 \frac{4}{5} \cdot \frac{39}{49} \cdot \frac{19}{24} = \frac{4 \cdot 39 \cdot 19}{5 \cdot 49 \cdot 24} = \frac{2964}{5880}

STEP 8

Let's simplify this big fraction by dividing the top and bottom by 12: 29645880=2964÷125880÷12=247490 \frac{2964}{5880} = \frac{2964 \div 12}{5880 \div 12} = \frac{247}{490} So, the probability of getting three working iPads in a row is 247490\frac{247}{490}.

STEP 9

Remember, we figured out the probability of *not* getting any broken iPads (i.e., getting all working iPads).
To find the probability of getting *at least one* broken iPad, we subtract the probability of getting *all working* iPads from **1** (because all probabilities add up to **1**).

STEP 10

1247490=490490247490=490247490=243490 1 - \frac{247}{490} = \frac{490}{490} - \frac{247}{490} = \frac{490 - 247}{490} = \frac{243}{490}

STEP 11

Now, let's convert this fraction to a decimal by dividing 243 by 490: 2434900.4959 \frac{243}{490} \approx 0.4959 Rounding to the nearest thousandth gives us **0.496**.

STEP 12

The probability of selecting at least one defective iPad is approximately **0.496**.

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