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jok3333 [9.3K]
3 years ago
9

Greg has a bag of beads that contains 16 black, 20 red, and 4 white beads. He randomly picks a head places it back into the bag

and then randomly picks another one. What is the probability that he picks a black and white bead?
Mathematics
1 answer:
jasenka [17]3 years ago
8 0

Answer:

4%

Step-by-step explanation:

There are 40 beads in total(16+20+4=40). There are 16 black so the probability of getting a black is 16/40=2/5. Since you put the bead back the bag will still have 40 beads when you pick another one. There are 4 white beads so probability of getting a white is 4/40=1/10. Then we multiply these two probabilities 2/5 * 1/10=1/25=4%.


Note: Order does not matter.

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Sauron [17]

Answer:

The expected monetary value of a single roll is $1.17.

Step-by-step explanation:

The sample space of rolling a die is:

S = {1, 2, 3, 4, 5 and 6}

The probability of rolling any of the six numbers is same, i.e.

P (1) = P (2) = P (3) = P (4) = P (5) = P (6) = \frac{1}{6}

The expected pay for rolling the numbers are as follows:

E (X = 1) = $3

E (X = 2) = $0

E (X = 3) = $0

E (X = 4) = $0

E (X = 5) = $0

E (X = 6) = $4

The expected value of an experiment is:

E(X)=\sum x\cdot P(X=x)

Compute the expected monetary value of a single roll as follows:

E(X)=\sum x\cdot P(X=x)\\=[E(X=1)\times \frac{1}{6}]+[E(X=2)\times \frac{1}{6}]+[E(X=3)\times \frac{1}{6}]\\+[E(X=4)\times \frac{1}{6}]+[E(X=5)\times \frac{1}{6}]+[E(X=6)\times \frac{1}{6}]\\=[3\times \frac{1}{6}]+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]\\+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]+[4\times \frac{1}{6}]\\=1.17

Thus, the expected monetary value of a single roll is $1.17.

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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Dennis_Churaev [7]

Answer:

The equation of the line is:

y=2x-1

Step-by-step explanation:

Given the points

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  • (1, 1)

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\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

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m=\frac{1-\left(-5\right)}{1-\left(-2\right)}

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