Question

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

Answer 1

First of all:
The number of ways to select 3 balls is: \[
^{3+4+5}P_3=^{12}P_3=\frac{12!}{(12-3)!} = \frac{12!}{9!}
\]

Answer 2

k

Answer 3

(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: \[
3!\times (3\times 4\times 5)
\]

Answer 4

do u mean 1320

Answer 5

The probability for one of each color is: \[
\frac{3!\times 3\times 4\times 5}{\frac{12!}{9!}}
\]

Answer 6

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.

Answer 7

Thus the probability of 3 red balls is: \[
\frac{3\times 2\times 1}{\frac{12!}{9!}}
\]

Answer 8

yes ur correct

Answer 9

The expected number of times that something happens \(n\) times with probability \(p\) is:\[
n\times p
\]
In this case \(n = 1100\).

Answer 10

Where is (c) and (d)?

Answer 11

Medal?

Answer 12

thanks

Answer 13

How did you know I was correct?

Answer 14

i have got answers at the back of my textbook but there wasn't any explanation