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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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Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
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  • p is the probability of a success on a single trial.

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  • A sample of 15 fatalities is taken, hence n = 15.

The probability is:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

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P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061

P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186

P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700

P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916

P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305

P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047

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P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

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<u>Explanation</u>

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Step-by-step explanation:

 

 

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