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ankoles [38]
1 year ago
15

A bag contains 20 identical balls of which 12 are white and 8 are black in colour. 3 balls are drawn randomly without any replac

ement. what is the probability of picking 2 white balls and 1 black ball?
Mathematics
1 answer:
Kaylis [27]1 year ago
7 0

The probability of picking 2 white balls and 1 black ball is 0.154

<h3>How to determine the probability?</h3>

The given parameters are:

  • Number of balls, n = 20
  • White = 12
  • Black = 8

When each ball is selected, the total number of ball decreases by 1 and the type of ball also decreases.

So, the probability is represented as:

P(2 white and 1 black) = White/n * White - 1/n - 1 * Black//n - 2

Substitute known values

P(2 white and 1 black) = 12/20 * 11/19 * 8//18

Evaluate

P(2 white and 1 black) = 0.154

Hence, the probability of picking 2 white balls and 1 black ball is 0.154

Read more about probability at:

brainly.com/question/251701

#SPJ1

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Kaitlyn is sharpening a pencil that is 19 cm. Each time she sharpens it goes down 2.5 cm. If she sharpens her pencil 4 times. Ho
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Answer:

9 cm

Step-by-step explanation:

2.5x4=10

7 0
3 years ago
Use the equation below to find c, if a = 43 and b = 49. c=180-a-) ?​
Luden [163]

Answer:

88

Step-by-step explanation:

43 + 49= 92.

Then 180 - 92= 88

C= 88

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3 years ago
Someone please help me i’m so llst
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4 0
2 years ago
The equatorial radius of Mars is 3,394 km. Which of the following would be a reasonable estimate for the equatorial radius of Ma
Ivan

Answer:

I don't know that options but......

if we are rounding to the thousands, then the answer would be 3,000

if hundreds, the answer is 3,400

if tens: 3,390

If ones: 3,390

Step-by-step explanation:

round each number up or down based upon what the digit is.

5 0
3 years ago
Read 2 more answers
If every student in a large Statistics class selects peanut M&amp;M’s at random until they get a red candy, on average how many
kotykmax [81]

Question:

According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M's from an extra-large bag looking for a red candy.

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.).

Answer:

If every student in the large Statistics class selects peanut M&M’s at random until they get a red candy the expected value of the number of  M&M’s the students need to select is 8.33 M&M's.

Step-by-step explanation:

To solve the question, we note that the statistical data presents a geometric mean. That is the probability of success is of the form.

The amount of repeated Bernoulli trials required before n eventual success outcome or

The probability of having a given number of failures before the first success is recorded.

In geometric distribution, the probability of having an eventual successful outcome depends on the the completion of a certain number of attempts with each having the same probability of success.

If the probability of each of the preceding trials is p and the kth trial is the  first successful trial, then the probability of having k is given by

Pr(X=k) = (1-p)^{k-1}p  

The number of expected independent trials to arrive at the first success for a variable Xis 1/p where p is the expected success of each trial hence p is the probability for the red and the expected value of the number of trials is 1/p or where p = 12 % which is 0.12

1/p = 1/0.12 or 25/3 or 8.33.

4 0
3 years ago
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