anonymous
  • anonymous
a bag contains 3 red 4 white and 5 green balls. Three balls are selected without replacement. Find the probability that the three balls chosen are: (a) one of each color If the selection of the three balls was carried out 1100 times, how often would you expect to choose: (b) 3 red balls (e) one of each color
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
First of all: The number of ways to select 3 balls is: \[ ^{3+4+5}P_3=^{12}P_3=\frac{12!}{(12-3)!} = \frac{12!}{9!} \]
anonymous
  • anonymous
k
anonymous
  • anonymous
(a) The number of ways to select one of each color, where order doesn't matter, is: \[ 3\times 4\times 5 \]Factoring in order (\(3!\) permutations) we get: \[ 3!\times (3\times 4\times 5) \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
do u mean 1320
anonymous
  • anonymous
The probability for one of each color is: \[ \frac{3!\times 3\times 4\times 5}{\frac{12!}{9!}} \]
anonymous
  • anonymous
For 3 red balls, the total number of ways is: \[ 3\times 2\times 1 \]There is only one possible permutation, so this alone is enough.
anonymous
  • anonymous
Thus the probability of 3 red balls is: \[ \frac{3\times 2\times 1}{\frac{12!}{9!}} \]
anonymous
  • anonymous
yes ur correct
anonymous
  • anonymous
The expected number of times that something happens \(n\) times with probability \(p\) is:\[ n\times p \] In this case \(n = 1100\).
anonymous
  • anonymous
Where is (c) and (d)?
anonymous
  • anonymous
Medal?
anonymous
  • anonymous
thanks
anonymous
  • anonymous
How did you know I was correct?
anonymous
  • anonymous
i have got answers at the back of my textbook but there wasn't any explanation

Looking for something else?

Not the answer you are looking for? Search for more explanations.