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vladimir2022 [97]
2 years ago
13

Henry has 3 black shirts and 7 blue shirts in his wardrobe. Two shirts are drawn without replacement from the wardrobe. What is

the probability that both of the shirts are black?
Mathematics
2 answers:
asambeis [7]2 years ago
6 0
Since there are three black shirts and 7 blue shirts, he has a 3/10 chance of picking a black.

Mars2501 [29]2 years ago
4 0

Answer:

The probability that both of the shirts are black is \frac{1}{15}

Step-by-step explanation:

Given : Henry has 3 black shirts and 7 blue shirts in his wardrobe. Two shirts are drawn without replacement from the wardrobe.

To find : What is the probability that both of the shirts are black?

Solution :

Number of black shirts = 3

Number of blue shirt = 7

Total number shirts = 3+7=10

The probability that first shirts is black is \frac{3}{10}

Two shirts are drawn without replacement from the wardrobe.

Number of black shirt left = 2

Total shirts = 9

The probability that second shirts is black is \frac{2}{9}

So, The probability that both of the shirts are black is

P=\frac{3}{10}\times\frac{2}{9}

P=\frac{6}{90}

P=\frac{1}{15}

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5 ( n - 7 ) = 2 ( n + 14 )
wariber [46]

Answer:

n=21

Step-by-step explanation:

We must find n.

Remember PEMDAS. First we must do the Parentheses. Lets do distributive property by multiplying a number that is immediately outside the parentheses with each number inside the parentheses. Lets do this one side at a time.

First lets do 5(n - 7). We get 5n - 35

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Now lets do OPPOSITES!!!! We must do the opposite of each thing to both sides.

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8 0
2 years ago
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If f(3)=2x^3 -2 , what is the value of f(2)
iris [78.8K]

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A game of chance involves rolling an unevenly balanced 4-sided die. The probability that a roll comes up 1 is 0.21, the probabil
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Answer:

2.65 dollars

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

E(x)=x1P(x1) + x2P(x2) + ... + xnP(xn)

Where x1, x2, ... , xn are the possibles values of the variable and P(x1), P(x2), ... , P(xn) are the probabilities of x1, x2, ... , xn respectively.

In this case, the roll can comes up 1, 2, 3 or 4 and you can win 1, 2, 3 or 4 dollars respectively. So, taking into account that they are mutually exclusive events, the probability that the player win 1, 2, 3 or 4 dollars is:

P(1) = 0.21

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P(3) = P(2∪3) - P(2) = 0.51 - 0.21 = 0.3

P(4) = 1 - P(1) - P(2) - P(3) = 1 - 0.21 - 0.21 - 0.3 = 0.28

Therefore, If you win the amount that appears on the die the expected winning are:

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E(x) = $1(0.21) + $2(0.21) + $3(0.3) + $4(0.28)

E(x) = $2.65

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