You and your mom enter a drawing with 3 different prizes. The prizes are
awarded at random.
If a total of 10 people entered the drawing, in how many ways can you win
first prize and your mom win second prize?
a. 8
b. 2
c. 10
d. 1

Answers

Answer 1

Answer:

The correct answer is 8.

Step-by-step explanation:

Answer 2

There are 2 ways that you can win first prize and your mom can win the second prize.

What is probability?

It is the chance of an event to occur from a total number of outcomes.

The formula for probability is given as:

Probability = Number of required events / Total number of outcomes.

Example:

The probability of getting a head in tossing a coin.

P(H) = 1/2

We have,

There are 10 people who entered the drawing.

Since the prizes are awarded at random, the probability of you winning the first prize is 1/3, and the probability of your mom winning the second prize is 1/2.

We can use the multiplication principle of counting to find the number of ways that you can win first prize and your mom can win the second prize.

The number of ways that you can win first prize is 1 since there is only one first prize and you are one of the 10 people who entered.

Once the first prize winner is chosen, there are 9 people left, and the number of ways that your mom can win the second prize is 1 out of 9.

The total number of ways that you can win first prize and your mom can win the second prize is:

1 × 1/3 × 1/2 × 9 = 3/2

However, the answer cannot be a fraction or decimal, since we are counting the number of ways in which events can occur.

Therefore, we round up to the nearest integer and get:

3/2 ≈ 2

Therefore,

There are 2 ways that you can win first prize and your mom can win the second prize.

Learn more about probability here:

https://brainly.com/question/14099682

#SPJ7


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Step-by-step explanation:

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========================================================

Work Shown:

[tex]\left(\frac{5x^3}{20xy^5}\right)^4\\\\\\\left(\frac{x^2}{4y^5}\right)^4\\\\\\\frac{\left(x^2\right)^4}{(4^1)^4\left(y^5\right)^4}\\\\\\\frac{x^{2*4}}{4^{1*4}y^{5*4}}\\\\\\\frac{x^{8}}{4^{4}y^{20}}\\\\\\\frac{x^{8}}{256y^{20}}\\\\\\[/tex]

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Step-by-step explanation:

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Step-by-step explanation:

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