Math  /  Data & Statistics

QuestionChapter 5 Section 5-1 and Multiplication Rule
1. 3 coins are tossed and a 6 -sided die is rolled. What is the probability of getting three tails and a multiple of 3 on the die?
2. What is the probability that a randomly selected number between 1 and 26 is divisible by 2 or 3 ?
3. In a bag of 100 stockings, 12 pairs have defects. Three dancers run into the dressing room and each grab a pair of stockings (without replacement). a) Find the probability that all three dancers get a pair of defective stockings. b) Find the probability that all three dancers get a pair on non-defective stockings. c) Find the probability that two dancers get non-defective stockings and the third dancer gets a defective pair.

Studdy Solution

STEP 1

What is this asking? We're figuring out the chances of getting specific outcomes when flipping coins, rolling dice, and grabbing stockings! Watch out! Don't mix up "and" and "or" probabilities, and remember that when we grab those stockings, we aren't putting them back!

STEP 2

1. Coin flips and die roll
2. Divisible numbers
3. Stockings

STEP 3

Let's **break down** this first problem.
We've got two independent events happening: flipping three coins and rolling a six-sided die.
We want to find the probability of getting three tails *and* a multiple of 3 on the die.

STEP 4

First, let's **tackle the coin flips**.
The probability of getting a tail on one coin flip is 12 \frac{1}{2} .
Since the coin flips are independent events, the probability of getting three tails in a row is 121212=18 \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} .

STEP 5

Now, let's **roll that die**!
The multiples of 3 on a six-sided die are 3 and 6.
So there are two favorable outcomes out of six possible outcomes.
The probability of rolling a multiple of 3 is 26=13 \frac{2}{6} = \frac{1}{3} .

STEP 6

Since the coin flips and the die roll are independent events, we **multiply their probabilities** to find the probability of *both* events happening.
That's 1813=124 \frac{1}{8} \cdot \frac{1}{3} = \frac{1}{24} .
Boom!

STEP 7

Alright, now we're **picking a number** between 1 and 26.
We want to know the probability that the number is divisible by 2 *or* 3.

STEP 8

Let's **find how many numbers** are divisible by 2.
Those are the even numbers: 2, 4, 6, ..., 26.
There are 262=13 \frac{26}{2} = 13 such numbers.

STEP 9

Next, let's **count the numbers** divisible by 3: 3, 6, 9, ..., 24.
There are 243=8 \frac{24}{3} = 8 such numbers.

STEP 10

Hold on!
Some numbers are divisible by *both* 2 and 3 (like 6, 12, 18, 24).
We've counted these numbers twice!
These are the multiples of 6, and there are 246=4 \frac{24}{6} = 4 of them.

STEP 11

So, the total number of numbers divisible by 2 *or* 3 is 13+84=17 13 + 8 - 4 = 17 .
The probability is then 1726 \frac{17}{26} .
Got it!

STEP 12

Time for some **stocking shenanigans**!
We've got 100 pairs of stockings, and 12 pairs are defective.
Three dancers each grab a pair without replacement.

STEP 13

**Part (a):** What's the probability that all three dancers get defective stockings?
The first dancer has a 12100 \frac{12}{100} chance.
The second dancer has an 1199 \frac{11}{99} chance (one less defective pair, one less total pair).
The third dancer has a 1098 \frac{10}{98} chance.
Multiplying these gives us 1210011991098=1320970200=118085 \frac{12}{100} \cdot \frac{11}{99} \cdot \frac{10}{98} = \frac{1320}{970200} = \frac{11}{8085} .

STEP 14

**Part (b):** Now, what's the probability they all get *non-defective* stockings?
There are 10012=88 100 - 12 = 88 non-defective pairs.
The probabilities are 88100 \frac{88}{100} , 8799 \frac{87}{99} , and 8698 \frac{86}{98} .
Multiplying gives us 8810087998698=658896970200=24313575 \frac{88}{100} \cdot \frac{87}{99} \cdot \frac{86}{98} = \frac{658896}{970200} = \frac{2431}{3575} .

STEP 15

**Part (c):** Two non-defective, one defective.
We can choose which dancer gets the defective pair in 3 ways.
The probability is 38810087991298=391776970200=275328970200=10343575 3 \cdot \frac{88}{100} \cdot \frac{87}{99} \cdot \frac{12}{98} = 3 \cdot \frac{91776}{970200} = \frac{275328}{970200} = \frac{1034}{3575} .

STEP 16

1. The probability of three tails and a multiple of 3 is 124 \frac{1}{24} .
2. The probability of a number between 1 and 26 being divisible by 2 or 3 is 1726 \frac{17}{26} .
3. a) The probability of all three dancers getting defective stockings is 118085 \frac{11}{8085} . b) The probability of all three dancers getting non-defective stockings is 24313575 \frac{2431}{3575} . c) The probability of two dancers getting non-defective stockings and one getting a defective pair is 10343575 \frac{1034}{3575} .

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